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Q.

For the cell reaction Zn(s)+Cu(aq)+2Zn(aq)+2+Cu(s), the decrease in gibbs free energy is -2.2FV. Find the magnitude of  electrical work done in one second for the cell reaction 2Zn(s)+2Cu(aq)2+2Znaq)+2+2 Cu(s).

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a

4.4FV

b

3.3FV

c

2.2FV

d

1.1FV

answer is A.

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Detailed Solution

Work done in one second is equal to the electrical potential multiplied by the total charge passed.

nFECell4×1.1=4.4FV

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