The number of electrons donated from substance(s) getting oxidized to the substance(s) getting reduced in the chemical equation for the following reaction is:
${Cr}_{2} O_{7}^{2 -} + {Fe}^{2 +} + C_{2} O_{4}^{2 -} \to {Cr}^{3 +} + {Fe}^{3 +} + {CO}_{2}$ (Unbalanced)

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TestHub · Accepted Answer

Writing the balanced oxidation and reduction half-reactions, making the number of electrons same in them and adding both half-reactions gives us the final balanced equation:

$[{Fe}^{2 +} + C_{2} O_{4}^{2 -} \to {Fe}^{3 +} + 2 {CO}_{2} + 3 e^{-}] \times 2 (oxidation)$

${Cr}_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 {Cr}^{3 +} + 7 H_{2} O (reduction)$

On adding both the reactions, we get,

${Cr}_{2} O_{7}^{2 -} + 2 {Fe}_{}^{2 +} + 2 C_{2} O_{4}^{2 -} + 14 H_{}^{+} \overset{}{\to} 2 {Cr}^{3 +} + 2 {Fe}^{3 +} + 4 {CO}_{2} + 7 H_{2} O$

Thus, $6$ electrons are donated from oxidized substances to the reduced substance.

Chemistry - Redox Reactions Question with Solution | TestHub

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