$a {MnO}_{4}^{-} + b I^{-} + c H^{+} ⟶ d {Mn}^{2 +} + e I_{2} + f H_{2} O$
In the above-balanced reaction, the value of $(\frac{c}{d})$ will be:

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1.2

1.3

answer is C.

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Detailed Solution

The given equation is :

$a {MnO}_{4}^{-} + b I^{-} + c H^{+} ⟶ d {Mn}^{2 +} + e I_{2} + f H_{2} O$

Reduction of half-reaction:

$8 H^{+} + {MnO}_{4}^{-} + 5 e^{-} \to {Mn}^{(} 2 +) + 4 H_{2} O\} \times 5$

Oxidation of half-reaction :

$2 I^{-} \to I_{1}^{0} + 2 e^{-}\} \times 2$

Then balance chemical equation is :

$2 {MnO}_{4}^{-} + 10 I^{-} + 16 H^{+} \to 2 {Mn}^{2 +} + 5 I_{2} + 8 H_{2} O$

$\begin{array}{l} c = 16 \\ d = 2 \\ \frac{c}{d} = \frac{16}{2} = 8 \end{array}$

Value of will be: 8

Hence, the correct option is C.

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