2 MnO_{4}^{-} + b C_{2}O_{4}^{2-} + c H^{+} r | vedclass

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(D) The given redox reaction is: $2 MnO_{4}^{-} + b C_{2}O_{4}^{2-} + c H^{+} \rightarrow x Mn^{2+} + y CO_{2} + z H_{2}O$Step $1$: Write the reductio

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$16$

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(D) The given redox reaction is: $2 MnO_{4}^{-} + b C_{2}O_{4}^{2-} + c H^{+} \rightarrow x Mn^{2+} + y CO_{2} + z H_{2}O$Step $1$: Write the reduction half-reaction:$MnO_{4}^{-} + 8 H^{+} + 5 e^{-} \rightarrow Mn^{2+} + 4 H_{2}O$Step $2$: Write the oxidation half-reaction:$C_{2}O_{4}^{2-} \rightarrow 2 CO_{2} + 2 e^{-}$Step $3$: Equalize the number of electrons by multiplying the reduction half-reaction by $2$ and the oxidation half-reaction by $5$:$2 MnO_{4}^{-} + 16 H^{+} + 10 e^{-} \rightarrow 2 Mn^{2+} + 8 H_{2}O$$5 C_{2}O_{4}^{2-} \rightarrow 10 CO_{2} + 10 e^{-}$Step $4$: Add the two half-reactions:$2 MnO_{4}^{-} + 5 C_{2}O_{4}^{2-} + 16 H^{+} \rightarrow 2 Mn^{2+} + 10 CO_{2} + 8 H_{2}O$Comparing this with the given equation,the coefficient $c$ corresponding to $H^{+}$ is $16$.

$2 MnO_{4}^{-} + b C_{2}O_{4}^{2-} + c H^{+} \rightarrow x Mn^{2+} + y CO_{2} + z H_{2}O$
If the above equation is balanced with integer coefficients,the value of $c$ is ...........
(Round off to the Nearest Integer).

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$2 MnO_{4}^{-} + b C_{2}O_{4}^{2-} + c H^{+} \rightarrow x Mn^{2+} + y CO_{2} + z H_{2}O$If the above equation is balanced with integer coefficients,the value of $c$ is ...........(Round off to the Nearest Integer).