The cell reaction of a cell is as follows: $M g (s) + C u_{(a q)}^{2 +} \to C u (s) + M g_{(a q)}^{2 +}$ . If the standard reduction potentials of $M g$ and $C u$ are $- 2.37$ V and $+ 0.34$ V respectively, the e.m.f. (in V) of the cell is:

+ 2.03

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- 2.03

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+ 2.71

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- 2.71

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Solution

The correct option is D $+ 2.71$
The cell reaction is as follows:

$M g (s) + C u^{2 +} (a q) \to C u (s) + M g^{2 +} (a q)$

In this reaction, ${C u}^{2 +}$ ion is reduced to copper metal while Mg is oxidized. The standard reduction potential values are given below:

$E_{C u} = + 0.34 V$ , $E_{M g} = - 2.37 V$

Hence, $E_{c e l l} = 0.34 - (- 2.37) = 2.71 V$

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The cell reaction of a cell is as follows: Mg(s)+Cu2+(aq)→Cu(s)+Mg2+(aq). If the standard reduction potentials of Mg and Cu are −2.37 V and +0.34 V respectively, the e.m.f. (in V) of the cell is:

The correct option is D +2.71The cell reaction is as follows:Mg(s)+Cu2+(aq)→Cu(s)+Mg2+(aq)In this reaction, Cu2+ ion is reduced to copper metal while Mg is oxidized. The standard reduction potential values are given below:ECu=+0.34 V, EMg=−2.37 VHence, Ecell=0.34−(−2.37)=2.71 V

The cell reaction of a cell is as follows: $M g (s) + C u_{(a q)}^{2 +} \to C u (s) + M g_{(a q)}^{2 +}$ . If the standard reduction potentials of $M g$ and $C u$ are $- 2.37$ V and $+ 0.34$ V respectively, the e.m.f. (in V) of the cell is: