\(E_{\text {cathode }}^{\circ}=(2.71-2.310) V=+0.40 \mathrm{V}\)

Short Answer

Expert verified
The reduction potential at the cathode is \(+0.40 V\).

Step by step solution

01

Substitute the known values into the equation

Given \(E_{\text {substance }}^{\circ}=2.71 V\) and \(E_{\text {reference }}^{\circ}=2.310 V\). Substitute these values into the equation: \(E_{\text {cathode }}^{\circ}=E_{\text {substance}}^{\circ}-E_{\text {reference}}^{\circ}\). This becomes: \(E_{\text {cathode }}^{\circ}=2.71 V - 2.310 V\)
02

Calculate the result

By implementing subtraction as per the previous step, this results in: \(E_{\text {cathode }}^{\circ}=+0.40 V\)

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