
In the above diagram, the standard electrode potentials are given in volts (over the arrow). The value of \( E^\circ_{\text{FeO}_4^{2-}/\text{Fe}^{2+}} \) is:
Consider the following electrochemical cell at standard condition. $$ \text{Au(s) | QH}_2\text{ | QH}_X(0.01 M) \, \text{| Ag(1M) | Ag(s) } \, E_{\text{cell}} = +0.4V $$ The couple QH/Q represents quinhydrone electrode, the half cell reaction is given below: $$ \text{QH}_2 \rightarrow \text{Q} + 2e^- + 2H^+ \, E^\circ_{\text{QH}/\text{Q}} = +0.7V $$
Assertion (A): Cu cannot liberate \( H_2 \) on reaction with dilute mineral acids.
Reason (R): Cu has positive electrode potential.
The elements of the 3d transition series are given as: Sc, Ti, V, Cr, Mn, Fe, Co, Ni, Cu, Zn. Answer the following:
Copper has an exceptionally positive \( E^\circ_{\text{M}^{2+}/\text{M}} \) value, why?

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