What is the cell reaction occurring in Daniell cell (galvanic cell)?

C u (s) + Z n S O_{4} (a q .) \to C u S O_{4} (a q .) + Z n (s)

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Z n (s) + C u S O_{4} (a q .) \to C u (s) + Z n S O_{4} (a q .)

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N i (s) + Z n S O_{4} (a q .) \to N i S O_{4} (a q .) + Z n (s)

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2 N a (s) + C d S O_{4} (a q .) \to N a_{2} S O_{4} (a q .) + C d (s)

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Solution

The correct option is B $Z n (s) + C u S O_{4} (a q .) \to C u (s) + Z n S O_{4} (a q .)$
In given Galvanic cell:-

$\to$ Standard electrode potential of $Z n^{2 +} / Z n$ is $- 0.76 V$ and of $C u^{2 +} / C u_{(s)}$ is $0.34 V$ .

$\to$ In the given reaction:-

$Z n_{(s)} + C u S O_{4 (a q)} \to C u_{(s)} + Z n S O_{4 (a q)}$ .

$Z n_{(s)}$ oxidizes to form $Z n S O_{4 (a q)}$ .

$C u S O_{4 (a q)}$ reduces to form $C u_{(s)}$ .

Suggest Corrections

Similar questions

Z n_{(} s) + C u S O_{4} (a q) \to Z n S O_{4} (a q) + C u_{(} s)

C u S O_{4 (a q)} + Z n_{(s)} \to Z n S O_{4 (a q)} + C u_{(s)}

(a) Z n_{(s)} + C u S o_{4 (a q)} \to Z n S O_{4 (a q)} + C u_{(s)}

(b) F e_{(s)} + Z n S O_{4 (a q)} \to F e S O_{4 (a q)} + Z n_{(s)}

(c) Z n_{(s)} + F e S O_{4 (a q)} \to Z n S O_{4 (a q)} + F e_{(s)}

+ 2 + 6 - 2 C u S O_{4 (a q)} + 0 Z n_{(s)} \to 0 C u_{(s)} + + 2 + 6 + - 2 Z n S O_{4 (a q)}

Z n (s) + C u S O_{4} (a q) \to Z n S O_{4} (a q) + C u (s)

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