A) 0.8 V
B) ? 0.8 V
C) ? 1.2 V
D) 1.2 V
Correct Answer: A
Solution :
\[\frac{1}{2}{{H}_{2}}|{{H}^{+}}|\] \[|A{{g}^{+}}|Ag|\] \[E_{Cell}^{0}=E_{Cathode}^{0}-E_{Anode}^{0}\]\[=E_{A{{g}^{+}}/Ag}^{0}-E_{{{H}^{+}}/\frac{1}{2}{{H}_{2}}}^{0}\] \[(0.80)-(0.0)=0.80\ V.\]
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