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If [Zn^(2+)] = 0.1 M and E^(@) = -0.76 ...

If `[Zn^(2+)] = 0.1 M ` and `E^(@) = -0.76 V` then half cell potential at 298 K for the reaction `Zn_((aq))^(2+)+2e^(-) rarr Zn_((s))` is

A

0.789 V

B

`-0.789V`

C

`-0.698V`

D

0.698 V

Text Solution

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The correct Answer is:
B
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