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Question

FInd Ecell for reduction of NO3NO by Cu(s), when [HNO3]=1M,PNO=103atm,[Cu2+]=0.1M
(Given: at 298 K RTF(2.303)=0.06volt;ENO3/NO=+0.96volt;ECu2+/Cu=+0.34volt)

A
0.61
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B
0.71
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C
0.51
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D
0.81
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Solution

The correct option is A 0.61
2NO3+3CuNO+Cu2+n=6Ecell=Eocell0.0591nlog(Cu2+)(NO3)=(0.960.34)0.05916log(0.11)Ecell=0.62+0.00985Ecell0.62

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