For the balanced redox reaction a C_2O_4^{2-} | vedclass

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Question

(A) Step $1$: Write the half-reactions.Oxidation: $C_2O_4^{2-} \rightarrow 2 CO_2 + 2 e^-$Reduction: $MnO_4^- + 8 H^+ + 5 e^- \rightarrow Mn^{2+} + 4

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$5, 2$

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(A) Step $1$: Write the half-reactions.Oxidation: $C_2O_4^{2-} \rightarrow 2 CO_2 + 2 e^-$Reduction: $MnO_4^- + 8 H^+ + 5 e^- \rightarrow Mn^{2+} + 4 H_2O$Step $2$: Balance the electrons by multiplying the oxidation half-reaction by $5$ and the reduction half-reaction by $2$.$5 C_2O_4^{2-} \rightarrow 10 CO_2 + 10 e^-$$2 MnO_4^- + 16 H^+ + 10 e^- \rightarrow 2 Mn^{2+} + 8 H_2O$Step $3$: Add the two equations to get the balanced net ionic equation:$5 C_2O_4^{2-} + 2 MnO_4^- + 16 H^+ \rightarrow 2 Mn^{2+} + 8 H_2O + 10 CO_2$Comparing this with the given equation, we get $a = 5$ and $x = 2$.

For the balanced redox reaction $a C_2O_4^{2-} + b MnO_4^- + c H^+ \rightarrow x Mn^{2+} + y H_2O + z CO_2$, the values of $a$ and $x$ respectively are:

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