Which of the following reactions cannot be a basis for electrochemical cell?

H_{2} + O_{2} \to H_{2} O

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A g N O_{3} + Z n \to Z n (N O_{3})_{2} + A g

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A g N O_{3} + N a C l \to A g C l ↓ + N a N O_{3}

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K M n O_{4} + F e S O_{4} + H_{2} S O_{4} \to K_{2} S O_{4} + F e_{2} (S O_{4})_{3} + M n S O_{4} + H_{2} O

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Solution

The correct option is C $A g N O_{3} + N a C l \to A g C l ↓ + N a N O_{3}$
Double displacement reaction:
$A g N O_{3}      (+ 1) + N a C l      (+ 1) ⟶ A g C l      (+ 1) ↓ + N a N O_{3}      (+ 1)$
Since, there is no redox reaction taking place hence no oxidation,
reduction therefore no electrochemical cell can be formed.

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K M n O_{4} + H 2 S O_{4} + F e S O 4 \to K 2 S O_{4} + M n S O_{4} + F e_{2} (S O_{4})_{3} + H_{2} O

F e C_{2} O_{4} + K M n O_{4} + H_{2} S O_{4} \to F e_{2} (S O_{4})_{3} + C O_{2} + M n S O_{4} + K_{2} S O_{4} + H_{2} O

A g N O_{3} + N a B r \to N a N O_{3} + A g B r

B a C l_{2} + H_{2} S O_{4} \to B a S O_{4} + 2 H C l

A g N O_{3} + N a C l \to A g C l + N a N O_{3}

N a O H + H C l \to N a C l + H_{2} O

K M n O_{4} + H_{2} S O_{4} + F e S O_{4} \to K_{2} S O_{4} + M n S O_{4} + F e_{2} (S O_{4})_{3} + H_{2} O

F e S O_{4} + K M n O_{4} + H_{2} S O_{4} \to F e_{2} (S O_{4})_{3} + M n S O_{4} + H_{2} O + K_{2} S O_{4}

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