# how to write half equations for redox reactions

Next the manganate(VII) half-equation is considered: $MnO_4^- \rightarrow Mn^{2+}\nonumber$. The $$\color{blue}{\textbf{oxidation half-reaction}}$$ is: $$\color{blue}{\textbf{Sn}^{2+}\textbf{(aq)} \to \textbf{Sn}^{4+}\textbf{(aq) + 2e}^{-}}$$. We multiply the reduction half-reaction by $$\text{2}$$ and the oxidation half-reaction by $$\text{5}$$ to balance the number of electrons in both equations: $$2\text{MnO}_{4}^{-}(\text{aq}) + 16\text{H}^{+}(\text{aq}) + 10\text{e}^{-}$$ $$\to$$ $$2\text{Mn}^{2+}(\text{aq}) + 8\text{H}_{2}\text{O}(\text{l})$$, $$5\text{H}_{2}\text{O}_{2}(\text{l})$$ $$\to$$ $$5\text{O}_{2}(\text{g}) + 10\text{H}^{+}(\text{aq}) + 10\text{e}^{-}$$, $$2\text{MnO}_{4}^{-}(\text{aq}) + 16\text{H}^{+}(\text{aq}) + 5\text{H}_{2}\text{O}_{2}(\text{l})$$ $$\to$$ $$5\text{O}_{2}(\text{g}) + 10\text{H}^{+}(\text{aq}) + 2\text{Mn}^{2+}(\text{aq}) + 8\text{H}_{2}\text{O}(\text{l})$$. Combining the half-reactions to make the ionic equation for the reaction. Adding two hydrogen ions to the right-hand side gives: $H_2O_2 \rightarrow O_2 + 2H^+\nonumber$. First, divide the equation into two halves; one will be an oxidation half-reaction and the other a reduction half- reaction, by grouping appropriate species. These two equations are described as "electron-half-equations" or "half-equations" or "ionic-half-equations" or "half-reactions" - lots of variations all meaning exactly the same thing! The product of the reduction of $$\text{H}_{2}\text{O}_{2}$$ in an alkaline medium is $$\text{OH}^{-}$$: $$\text{H}_{2}\text{O}_{2}(\text{l})$$ $$\to$$ $$\text{OH}^{-}(\text{aq})$$, $$\text{H}_{2}\text{O}_{2}(\text{l})$$ $$\to$$ $$2\text{OH}^{-}(\text{aq})$$, $$\text{H}_{2}\text{O}_{2}(\text{l}) + 2\text{e}^{-}$$ $$\to$$ $$2\text{OH}^{-}(\text{aq})$$. There are three definitions you can use for oxidation: 1. $$\color{blue}{\textbf{Tin}}$$ is therefore being $$\color{blue}{\textbf{oxidised}}$$ and $$\color{red}{\textbf{iron}}$$ is the $$\color{red}{\textbf{oxidising agent}}$$ (causing tin to be oxidised). There are 3 positive charges on the right-hand side, but only 2 on the left. The LibreTexts libraries are Powered by MindTouch® and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. The unbalanced reduction half-reaction is: $$\text{Cl}_{2}(\text{g})$$ $$\to$$ $$\text{Cl}^{-}(\text{aq})$$. Fe 2+ + Cr → Fe + Cr 3+ Solution. Six electrons are added to the left to give a net +6 charge on each side. Write the half reactions. The chlorine reaction, in which chlorine gas is reduced to chloride ions, is considered first: The atoms in the equation must be balanced: This step is crucial. We check the number of atoms and the charges and find that the equation is balanced. The atoms balance, however the charges do not. And one way to check that your half reactions actually makes sense is you can actually sum up the two sides. $$\color{red}{\textbf{Iron}}$$ is therefore being $$\color{red}{\textbf{reduced}}$$ and $$\color{blue}{\textbf{tin}}$$ is the $$\color{blue}{\textbf{reducing agent}}$$ (causing iron to be reduced). Don't worry if it seems to take you a long time in the early stages. Two electrons must be added to the left hand side to balance the charges. reduction reaction: Y 2+ (aq) + 2e-→ Y (s) oxidation reaction: X (s) → X + (aq) + e-Step 2: Use tabulated values to find the standard electrode potential for each half-equation: The two balanced half reactions are summarized: The least common multiple of 4 and 6 is 12. Hydrogen ions are a better choice. Add water molecules to the left and $$\text{H}^{+}$$ ions to the right to balance the oxygen and hydrogen atoms: $$\text{S}^{2-}(\text{aq}) + 4\text{H}_{2}\text{O}(\text{l})$$ $$\to$$ $$\text{SO}_{4}^{2-}(\text{aq}) + 8\text{H}^{+}(\text{aq})$$. Half-reactions can be used to balance redox reactions. In the process, the chlorine is reduced to chloride ions. This makes sense as electrons are gained in the reduction half-reaction. Balance the charge by adding an electron to the left (this makes sense as this is the reduction half-reaction, and $$\text{N}^{5+}$$ $$\to$$ $$\text{N}^{4+}$$): $$\text{NO}_{3}^{-}(\text{aq}) + 2\text{H}^{+}(\text{aq}) + \text{e}^{-}$$ $$\to$$ $$\text{NO}_{2}(\text{g}) + \text{H}_{2}\text{O}(\text{l})$$, $$\text{S}^{2-}(\text{aq})$$ $$\to$$ $$\text{SO}_{4}^{2-}(\text{aq})$$. Redox equations where four half-reactions are required. As some curricula do not include this type of problem, the process for balancing alkaline redox reactions is covered on a separate page. Four hydrogen ions to the right-hand side to balance the hydrogen atoms: $CH_3CH_2OH + H_2O \rightarrow CH_3COOH + 4H^+\nonumber$. Reminder: a redox half-reaction MUST be balanced both for atoms and charge in order to be correct. The fully balanced half-reaction is: by adding seven water molecules to the right: Working out electron-half-equations and using them to build ionic equations, Balancing reactions under alkaline conditions, information contact us at info@libretexts.org, status page at https://status.libretexts.org, hydrogen ions (unless the reaction is being done under alkaline conditions, in which case, hydroxide ions must be added and balanced with water). Two electrons must be added to the right hand side of the equation. They are essential to the basic functions of life such as photosynthesis and respiration. The ionic equation for the magnesium-aided reduction of hot copper(II) oxide to elemental copper is given below : $Cu^{2+} + Mg \rightarrow Cu + Mg^{2+}\nonumber$. oxidation half-reaction: $$\color{red}{\times \textbf{2}}$$: $$\color{red}{2}$$$$\text{Fe}^{2+}(\text{aq})$$ $$\to$$ $$\color{red}{2}$$$$\text{Fe}^{3+}(\text{aq}) +$$$$\color{red}{\textbf{2}}{\textbf{e}^{-}}$$, reduction half-reaction: $$\color{red}{\times \textbf{1}}$$: $$\text{Cl}_{2}(\text{g}) +$$ $$\textbf{2e}^{-} \to$$ $$2\text{Cl}^{-}(\text{aq})$$, $$2\text{Fe}^{2+}(\text{aq}) + \text{Cl}_{2}(\text{g})$$ $$\to$$ $$2\text{Fe}^{3+}(\text{aq}) + 2\text{Cl}^{-}(\text{aq})$$. Subtracting 10 hydrogen ions from both sides leaves the simplified ionic equation. Write balance equations for the following redox reactions: a. NaBr + Cl 2 NaCl + Br 2 b. Fe 2 O 3 + CO Fe + CO 2 in acidic solution c. CO + I 2 O 5 CO 2 + I 2 in basic solution Hint; Write balanced equations for the following reactions: Hint. However, $$\text{H}_{2}\text{S}(\text{g})$$ is the reactant, so it would be better to write: $$\text{Cr}_{2}\text{O}_{7}^{2-}(\text{aq}) + 8\text{H}^{+}(\text{aq}) + 3\text{H}_{2}\text{S}(\text{g})$$ $$\to$$ $$3\text{S}(\text{s}) + 2\text{Cr}^{3+}(\text{aq}) + 7\text{H}_{2}\text{O}(\text{l})$$. The oxidation numbers of some elements must increase, and others must decrease as reactants go to products. Legal. $$\text{Cr}_{2}\text{O}_{7}^{2-}(\text{aq}) + 14\text{H}^{+}(\text{aq})$$ $$\to$$ $$2\text{Cr}^{3+}(\text{aq}) + 7\text{H}_{2}\text{O}(\text{l})$$. Embedded videos, simulations and presentations from external sources are not necessarily covered Every redox reaction is made up of two half-reactions: in one, electrons are lost (an oxidation process); in the other, those electrons are gained (a reduction process). Permanganate(VII) ions ( $$\text{MnO}_{4}^{-}$$ ) oxidise hydrogen peroxide ( $$\text{H}_{2}\text{O}_{2}$$ ) to oxygen gas. Balancing in a basic solution follows the same steps as above, … easily resolved by adding two electrons to the left-hand side. At this stage, students often forget to balance the chromium atoms, making it impossible to obtain the overall equation. In $$\text{Cr}_{2}\text{O}_{7}^{2-}$$ chromium exists as $$\text{Cr}^{6+}$$. There is a net +7 charge on the left-hand side (1- and 8+), but only a charge of +2 on the right. In this case, the least common multiple of electrons is ten: The equation is not fully balanced at this point. The reaction is carried out with potassium manganate(VII) solution and hydrogen peroxide solution acidified with dilute sulfuric acid. Each half-reaction is then balanced individually, and then the half-reactions are added back together to form a new, balanced redox equation. Split reaction into two half-reactions. Balance the atoms apart from oxygen and hydrogen. This illustrates the strategy for balancing half-equations, summarized as followed: Now the half-equations are combined to make the ionic equation for the reaction. a. Cr(OH) 3 + Br 2 CrO 4 2-+ Br-in basic solution. To remember this, think that LEO the lion says GER (Loss of Electrons is Oxidation; Gain of Electrons is Reduction). The oxygen atoms are balanced by adding seven water molecules to the right: $Cr_2O_7^{2-} \rightarrow 2Cr^{3+} + 7H_2O\nonumber$. Adding the two equations together and removing the electrons gives: $$2\text{I}^{-}(\text{aq}) + 2\text{Fe}^{3+}(\text{aq})$$ $$\to$$ $$\text{I}_{2}(\text{s}) + 2\text{Fe}^{2+}(\text{aq})$$. Therefore, in $$\text{Co}(\text{NH}_{3})_{6}^{2+}$$ cobalt exists as $$\text{Co}^{2+}$$. Something is being oxidized. From this information, the overall reaction can be obtained. This technique can be used just as well in examples involving organic chemicals. Removing any extra $$\text{H}^{+}$$ ions we get: $$2\text{MnO}_{4}^{-}(\text{aq}) + 6\text{H}^{+}(\text{aq}) + 5\text{H}_{2}\text{O}_{2}(\text{l})$$ $$\to$$ $$5\text{O}_{2}(\text{g}) + 2\text{Mn}^{2+}(\text{aq}) + 8\text{H}_{2}\text{O}(\text{l})$$. DON'T FORGET TO CHECK THE CHARGE. In an acid medium there are water molecules and $$\text{H}^{+}$$ ions in the solution, so these can be used to balance the equation. Is there a compound or atom being reduced? Putting the spectator ions back into the equation we get: $$2\text{NaI}(\text{aq}) + \text{Fe}_{2}(\text{SO}_{4})_{3}(\text{aq})$$ $$\to$$ $$\text{I}_{2}(\text{s}) + 2\text{FeSO}_{4}(\text{aq}) + \text{Na}_{2}\text{SO}_{4}$$. 1525057, and 1413739 3+ solution or reduced, and on any device oxidation and reduction give. Two half-reactions, one showing the reduction potentials of two half-reactions, one showing oxidation. Same number of electrons is Ten: the least common multiple of 4 and 6 12! 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