Relative efficiency of the following cell is $84$ %
$A_{(s)} + B_{(a q)}^{2 +} ⇌ A_{(a q)}^{2 +} + B_{(A)}, Δ G = - 285 k J$
Then the standard electrode potential of the cell will be:

1.24 V

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2.40 V

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1.10 V

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1.28 V

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Solution

The correct option is B $1.24 V$
$A (s) + B^{2 +} (a q) ⇌ A^{2 +} (a q) + B$
$x = 2$
$Δ G = - 285 k J$
$\Rightarrow - 285 \times 10^{3} = - 2 \times 96485 \times E$
$\Rightarrow E = 1.47 V$
$∴$ Standard electrode potential $= 1.47 \times 0.84 = 1.24 V$

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Relative efficiency of the following cell is 84%A(s)+B2+(aq)⇌A2+(aq)+B(A),ΔG=−285kJThen the standard electrode potential of the cell will be:

The correct option is B 1.24VA(s)+B2+(aq)⇌A2+(aq)+Bx=2ΔG=−285kJ⇒−285×103=−2×96485×E⇒E=1.47V∴ Standard electrode potential =1.47×0.84=1.24V

Relative efficiency of the following cell is $84$ %
$A_{(s)} + B_{(a q)}^{2 +} ⇌ A_{(a q)}^{2 +} + B_{(A)}, Δ G = - 285 k J$
Then the standard electrode potential of the cell will be:

The correct option is B $1.24 V$
$A (s) + B^{2 +} (a q) ⇌ A^{2 +} (a q) + B$
$x = 2$
$Δ G = - 285 k J$
$\Rightarrow - 285 \times 10^{3} = - 2 \times 96485 \times E$
$\Rightarrow E = 1.47 V$
$∴$ Standard electrode potential $= 1.47 \times 0.84 = 1.24 V$