Question

$\text{Ni/N}{{\text{i}}^{\text{2+}}}\text{//C}{{\text{u}}^{\text{2+}}}\text{/Cu}$
The standard EMF of the above cell is 0.59 V. The standard electrode potential (reduction potential) of the copper electrode is 0.34 V. Calculate the standard electrode potential of nickel electrode.
(A) -0.5 V
(B) 0.25 V
(C) 0.93 V
(D) -0.25 V

Answer 1

Find the half where reduction and oxidation is taking place.
We could solve this problem using the equation:$E_{cell}^{0}=E_{cathode}^{0}-E_{anode}^{0}$

Complete step by step answer:
- So let’s analyze the question and find the requirements of the question. It is given that the standard EMF of the given cell is 0.59V and the standard electrode potential of the Cu electrode is given as 0.34V.It is also given that the potential given is the reduction potential. From the information given we have calculated the standard electrode of the nickel electrode.
- So now we know the given data and requirement for the question.
Now we have to relate these terms and solve the problem using an appropriate equation.
First analyze the cell reaction given.
In the cell the pure Ni is converted to $\text{N}{{\text{i}}^{\text{2+}}}$, here the oxidation number of the element is increasing and we could say that Ni is getting oxidized. And the cell in which oxidation takes place is the anode cell.
- Now consider the reaction of Cu. In the cell $\text{C}{{\text{u}}^{\text{2+}}}$ is getting reduced to Cu. So the reduction of $\text{C}{{\text{u}}^{\text{2+}}}$ is taking place and the reduction takes place in a cathode. The oxidation number is getting reduced from +2 to 0.
Since we know that the standard EMF of the whole cell i.e. $E_{cell}^{0}$ is 0.59 V and we also know the standard EMF of the cathode which is the potential of Cu electrode. Relate these two values with the standard potential of Ni electrode.
We know the equation that:
$\text{E}_{\text{cell}}^{\text{0}}\text{=E}_{\text{cathode}}^{\text{0}}\text{-E}_{\text{anode}}^{\text{0}}$
$\text{E}_{\text{cell}}^{\text{0}}\text{=E}_{\text{reduction}}^{\text{0}}\text{-E}_{\text{oxidation}}^{\text{0}}$
Substitute the values in the given equation.
$\text{0}\text{.59=0}\text{.34-E}_{\text{oxidation}}^{\text{0}}$
- Write the equation for \text{E}_{\text{oxidation}}^{\text{0}}$$$$\left( \text{E}_{\text{Ni cell}}^{\text{0}} \right)
$\text{-E}_{\text{oxidation}}^{\text{0}}\text{=0}\text{.59-0}\text{.34=0}\text{.25V}$
$\text{E}_{\text{oxidation}}^{\text{0}}\text{=-0}\text{.25V}$
- Therefore,the standard electrode potential of Ni electrode is ,$\text{E}_{\text{Ni cell}}^{\text{0}}\text{=-0}\text{.25V}$ The correct answer is option “D” .

Note: We could remember a code to avoid confusion in which cell the oxidation and reduction takes place.
AN OX which represents Anode-oxidation, oxidation will take place in the anode of the cell.
RED CAT- reduction-cathode, which means reduction takes place in cathode.
So we could combine and memorize as RED CAT-AN OX.
- If in the question concentration is given then we use the Nernst equation which is,
${{\text{E}}_{\text{cell}}}\text{=E}_{\text{cell}}^{\text{0}}\text{-}\dfrac{\text{0}\text{.0591}}{\text{2}}\text{log}\dfrac{\left[ \text{Reduction} \right]}{\left[ \text{Oxidation} \right]}$