How to find standard electrode potential of a reaction resulting from two different reactions?

Question

Q1) Calculate $E_{\ce{Cu+}|\ce{Cu}}^o=E_o$. Given that- $E_{\ce{Cu^2+}|\ce{Cu}}^o=E_1$ and $E_{\ce{Cu+}|\ce{Cu^2+}}^o=E_2$

Method 1- $$(i)\quad\ce{Cu^2+} + 2e^-\to \ce{Cu} \quad... \Delta G_1=-2FE_1$$ $$(ii)\quad\ce{Cu+}\to\ce{Cu^2+}+e^-\quad...\Delta G_2=-FE_2$$

Required reaction is the "addition" of these two so- $$\ce{Cu+}+e^-\to\ce{Cu}$$ $$E_o=2E_1+E_2$$ Method 2- Just directly add the two reactions, which gives us $E_o=E_1+E_2$

Q2) Calculate standard cell potential of the following reaction- $$14\ce{H+}+6\ce{Cl-}+\ce{Cr2O7^2-}\to 3\ce{Cl2}+2\ce{Cr^3+}+7\ce{H2O}$$ Given that $E_{\ce{Cr2O7^2-}|\ce{Cr^3+}}^o=E_1$ and $E_{\ce{Cl-}|\ce{Cl2}}^o=E_2$

Method 1- Use the same process as done in the above example i.e. write expressions of $\Delta G$ of reduction of dichromate and oxidation of chloride, multiply them by suitable coefficients so as to get the desired chemical equation. This gives us $E_o=\frac{E_1+3E_2}{6}$.

Method 2- Just directly add the two reactions, which gives us $E_o=E_1+E_2$

According to my textbook Method 1 is correct for the first question while Method 2 is correct for the second. Why?

Answer 1

Q2) Calculate standard cell potential of the following reaction- $$14\ce{H+}+6\ce{Cl-}+\ce{Cr2O7^2-}\to 3\ce{Cl2}+2\ce{Cr^3+}+7\ce{H2O}$$ Given that $E_{\ce{Cr2O7^2-}|\ce{Cr^3+}}^o=E_1$ and $E_{\ce{Cl-}|\ce{Cl2}}^o=E_2$

The half-cell reactions are:$$14\ce{H+}+6\ce{e-}+\ce{Cr2O7^2-}\to 2\ce{Cr^3+}+7\ce{H2O};~~~~~~~~~~n_1=6$$ $$3[\ce{2Cl-}\to \ce{Cl2}+\ce{2e-}];~~~~~~~~~~n_2=6$$

Note that, before arriving the final equation, you have to balance the charges and then cancel out the electrons by multiplying with appropriate coefficient. As $G$ is a state function: $$\Delta G^0_1+\Delta G^0_2=\Delta G^0$$ $$\implies -n_1FE^0_1+(-n_2FE^0_2)=-nFE^0$$ $$\implies E^o=\frac{6E^0_1+6E^0_2}{6}=E^0_1+E^0_2$$

Q1) Calculate $E_{\ce{Cu+}|\ce{Cu}}^o=E_o$. Given that- $E_{\ce{Cu^2+}|\ce{Cu}}^o=E_1$ and $E_{\ce{Cu+}|\ce{Cu^2+}}^o=E_2$

Similarly, the half-cell reactions are: $$(i)\quad\ce{Cu^2+} + 2e^-\to \ce{Cu} \quad... \Delta G^0_1=-2FE^0_1$$ $$(ii)\quad\ce{Cu+}\to\ce{Cu^2+}+e^-\quad...\Delta G^0_2=-FE^0_2$$

Required reaction is the "addition" of these two so- $$\ce{Cu+}+e^-\to\ce{Cu};~~~~~~~~~~n=1$$ $$\Delta G^0_1+\Delta G^0_2=\Delta G^0$$ $$\implies -2FE^0_1+(-FE^0_2)=-FE^0$$ $$\implies E^o=\frac{2E^0_1+E^0_2}{1}=2E^0_1+E^0_2$$

Therefore, we can always get the correct result by equating the values of $\Delta G$.