Given: E∘Ag+∣Ag=0.799V and E∘Zn2+∣Zn=−0.673V, which of the following is true?

# Given: ${E}^{\circ }\left({\mathrm{Ag}}^{+}\mid \mathrm{Ag}\right)=0.799\mathrm{V}$ and ${E}^{\circ }\left({\mathrm{Zn}}^{2+}\mid \mathrm{Zn}\right)=-0.673\mathrm{V}$, which of the following is true?

1. A

${\mathrm{Ag}}^{+}$can be reduced by ${\mathrm{H}}_{2}\left(\mathrm{g}\right)$

2. B

Ag can be oxidized by ${\mathrm{H}}^{+}$

3. C

${\mathrm{Zn}}^{2+}$ can be reduced by ${\mathrm{H}}_{2}\left(\mathrm{g}\right)$

4. D

Ag can reduce ${\mathrm{Zn}}^{2+}$

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### Solution:

Given: ${E}^{\circ }\left({\mathrm{Ag}}^{+}\mid \mathrm{Ag}\right)=0.799\mathrm{V}$ and ${E}^{\circ }\left({\mathrm{Zn}}^{2+}\mid \mathrm{Zn}\right)=-0.673\mathrm{V}$

From the above equation it is observed that  more positive potential will constitute positive terminal (i.e. cathode) and the ions will be reduced here. So ${\mathrm{Ag}}^{+}$ is reduced by ${\mathrm{H}}_{2}$ gas.

Hence, ${\mathrm{Ag}}^{+}$can be reduced by ${\mathrm{H}}_{2}$

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