Calculate the E and E- of the cell Ni - Ni2- - - Cu2- - Cufrom the following half-cell reactions- Ni2- - 2e- Ni- E- - -0-25 volt Cu2- - 2e- Cu- E- - - 0-34 voltGiven- -Ni2- - 1 M and -Cu2- - 10-3M-

E^∘_cell=0.34+035=0.59V E_cell=E^∘_cell RTnFlnQ=0.59 (8.314)(298)(2)(96500)ln[Ni^2+][Cu^2+] Cu^2++Ni⟶ Ni^2++Cu =0.59 (0.0128)ln(110 3) ...

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Question

Calculate the $E$ and $E^{\circ}$ of the cell
$N i | N i^{2 +} | | C u^{2 +} | C u$
from the following half-cell reactions:
$N i^{2 +} + 2 e^{-} \to N i; E^{\circ} = - 0.25 v o l t$
$C u^{2 +} + 2 e^{-} \to C u; E^{\circ} = + 0.34 v o l t$
(Given: $[N i^{2 +}] = 1 M$ and $[C u^{2 +}] = 10^{- 3} M)$ .

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