    Question

# Calculate the emf of the following cell at 298 K. Zn(s)∣Zn2+(aq, 0.1 M)∥Ag+(aq, 0.01 M)∣ Ag(s) Given: E0Zn2+/Zn = −0.76 V E0Ag+/Ag =+0.80 V

A
Ecell=1.472 V
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B
Ecell=1.472 V
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C
Ecell=+1.732 V
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D
Ecell=1.732 V
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Solution

## The correct option is A Ecell=1.472 VFrom the given cell representation, Ag+/Ag couple act as cathode Zn2+/Zn couple act as anode E0cell=E0cathode−E0anode E0cell=0.80 − (−0.76) E0cell=1.56 V The given cell reaction is, Zn(s)+2Ag+(aq)→Zn2+(aq)+2Ag n=2 By Nernst equation, Ecell = Eocell − 0.0591nlog[Zn2+][Ag+]2 Ecell = Eocell + 0.0591nlog[Ag+]2[Zn2+] Ecell= 1.56 + 0.05912log[0.01]2[0.1] Ecell= 1.56 + 0.05912 log (1 ×10−3) Ecell= 1.56 − 0.05912×3 Ecell= 1.56 − 0.088 = 1.472 V  Suggest Corrections  5      Similar questions  Explore more