EMF of cell NiNi2+(1.0M)∥Au3+(1.0M)Au is......., if E∘ for Ni2+∣Ni is −0.25V,E∘ for Au3+∣Au is 1.50V.

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+1.255V

-1.75 V

+1.75 V

+4.0 V

answer is C.

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Detailed Solution

Ecell=EOPNi/Ni22+∘+ERPAu3+/Au∘EORNi∘−0.0592log⁡Ni2++ERPAu∘+log⁡0.0593Au3+=0.25−0.0592log⁡(1.0)+1.50+0.0593log⁡1.0=1.75V

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