The reduction potential for the given electrodes can be calculated using the Nernst equation: \[ E = E^\circ - \frac{0.0591}{n} \log \left( \frac{[\text{Reductant}]}{[\text{Oxidant}]} \right) \] Since the number of electrons involved in the silver (Ag) reduction reaction is 1 (i.e., \( n = 1 \)), the equation simplifies to: \[ E = E^\circ - 0.0591 \log [\text{Ag}^+] \] The standard reduction potential \( E^\circ_{\text{Ag}^+/\text{Ag}} \) is given as +0.80 V for standard conditions.
Using the Nernst equation: - For electrode A: \[ E_A = 0.80 - 0.0591 \log (0.0001) = 0.80 - 0.0591 \times (-4) = 0.80 + 0.2364 = 1.0364 \text{ V} \] - For electrode B: \[ E_B = 0.80 - 0.0591 \log (0.1) = 0.80 - 0.0591 \times (-1) = 0.80 + 0.0591 = 0.8591 \text{ V} \] - For electrode C: \[ E_C = 0.80 - 0.0591 \log (0.01) = 0.80 - 0.0591 \times (-2) = 0.80 + 0.1182 = 0.9182 \text{ V} \] - For electrode D: \[ E_D = 0.80 - 0.0591 \log (0.001) = 0.80 - 0.0591 \times (-3) = 0.80 + 0.1773 = 0.9773 \text{ V} \]
The reduction potentials for the electrodes in descending order are: \[ \text{B} > \text{C} > \text{D} > \text{A} \] Therefore, the correct order is Option (A): B > C > D > A.
Correct Answer: Option (A)
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:
The elements of the 3d transition series are given as: Sc, Ti, V, Cr, Mn, Fe, Co, Ni, Cu, Zn. Answer the following:
(a) Copper has an exceptionally positive \( E^\circ_{\text{M}^{2+}/\text{M}} \) value, why?