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Equilibrium constant and standard EMF of a cell (log K = nE/0.059)

This type of question, asked 3 times: 29 Jan 2023, 31 Jan 2023, 21 Jan 2026 S2. All 3 questions below, oldest first.

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  1. Question 1

    29 Jan 2023

    144
    The equilibrium constant for the reaction Zn(s)+Sn2+(aq)⇌Zn2+(aq)+Sn(s)Zn(s) +Sn^{2 +}(aq) \rightleftharpoons Zn^{2 +}(aq) + Sn(s) is 1×10201 \times10^{20} at 298 K298\text{ }K. The magnitude of standard electrode potential of Sn/Sn2+Sn/Sn^{2 +} if EZn2+/Zn∘=−0.76 VE_{Zn^{2 +}/Zn}^{\circ}= - 0.76\text{ }V is ____×10−2 V\_\_\_\_\times10^{- 2}\text{ }V (Nearest integer). Given: 2.303RTF=0.059 V\frac{2.303RT}{F}= 0.059\text{ }V

  2. Question 2

    31 Jan 2023

    53
    The logarithm of equilibrium constant for the reaction Pd2++4Cl−⇌PdCl42−Pd^{2 +}+ 4Cl^{-}\rightleftharpoons PdCl_{4}^{2 -} is (Nearest integer)
     Given : 2.303RTF=0.06 VPd(aq)2++2e−⇌Pd(s)E⊖=0.83 VPdCl42−(aq)+2e−⇌Pd(s)+4Cl−(aq)Eθ=0.65 V\begin{matrix} & \text{~Given :~}\frac{2.303RT}{F}= 0.06\text{ }V \\ & Pd_{(aq)}^{2 +}+ 2e^{-}\rightleftharpoons Pd(s)E^{\ominus}= 0.83\text{ }V \\ & PdCl_{4}^{2 -}(aq) + 2e^{-}\rightleftharpoons Pd(s) + 4Cl^{-}(aq)E^{\theta}= 0.65\text{ }V \end{matrix}

  3. Question 3

    21 Jan 2026 S2

    50
    MX is a sparingly soluble salt that follows the given solubility equilibrium at 298 K
    MX(s)⇌M+(aq)+X−(aq);Ksp=10−10MX(s) \rightleftharpoons M^{+}(aq) + X^{-}(aq);K_{sp} = 10^{- 10}
    If the standard reduction potential for M+(aq)→+e− M(s)M^{+}(aq)\overset{+ e^{-}\ }{\rightarrow}M(s) is (EM+/M⊖)=0.79 V\left( E_{M^{+}/M}^{\ominus} \right) = 0.79\text{ }V, then the value of the standard reduction potential for the metal/metal insoluble salt electrode EX−/MX(s)/M⊖E_{X^{-}/MX(s)/M}^{\ominus} is ____\_\_\_\_ mV. (nearest integer) [Given: 2.303RTF=0.059 V\frac{2.303RT}{F} = 0.059\text{ }V ]