Which of the following is the correct cell representation for the given cell reaction?
$Z n + H_{2} S O_{4} \to Z n S O_{4} + H_{2}$

Z n | Z n^{2 +} | | H^{+} | H_{2}

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Z n | Z n^{2 +} | | H^{+}, H_{2} | P t

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Z n | Z n S O_{4} | | H_{2} S O_{4} | Z n

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Z n | H_{2} S O_{4} | | Z n S O_{4} | H_{2}

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Solution

The correct option is B $Z n | Z n^{2 +} | | H^{+}, H_{2} | P t$
In the given reaction the oxidation state of $Z n$ is changing from 0 to +2 so it is undergoing oxidation. Whereas the oxidation state of $H^{+}$ is changing from +1 to 0 which implies it is undergoing reduction. And in cell representation oxidation is represented in left part and the reduction in right. It has to be remembered that $H_{2}$ electrode has Pt too in it.
So, the correct representation of the cell is,
$Z n | Z n^{2 +} | | H^{+}, H_{2} | P t$
Hence option (B) is correct

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Similar questions

Z n + H_{2} S O_{4} \to Z n S O_{4} + H_{2}

Z n + H_{2} S O_{4} \to Z n S O_{4} + H_{2}

What information does the following chemical equations convey ?

(a) $Z n$ + $H_{2} S O_{4}$ $\to$ $Z n S O_{4}$ + $H_{2}$

(b) $M g$ + $2 H C l$ $\to$ $M g C l_{2}$ + $H_{2}$

Z n + H_{2} S O_{4} (d i l) \to Z n S O_{4} + H_{2} ↑

Z n + d i l . H_{2} S O_{4} \to Z n S O_{4} + H_{2}

Z n + c o n c . H_{2} S O_{4} \to Z n S O_{4} + S O_{2} + H_{2} O

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