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Potential of a half-cell that depends on pH (Nernst equation)

This type of question, asked 3 times: 24 Jan 2023, 1 Feb 2023, 22 Jan 2026 S2. All 3 questions below, oldest first.

Try each one first, then open its answer.

  1. Question 1

    24 Jan 2023

    60
    At 298 K298\text{ }K, a 1 litre solution containing 10mmol10mmol of Cr2O72−Cr_{2}O_{7}^{2 -} and 100mmol100mmol of Cr3+Cr^{3 +} shows a pH of 3.0. Given: Cr2O72−→Cr3+;E∘=1.330 VCr_{2}O_{7}^{2 -}\rightarrow Cr^{3 +};E^{\circ}= 1.330\text{ }V and 2.303RTF=0.059 V\frac{2.303RT}{F}= 0.059\text{ }V The potential for the half cell reaction is x×10−3 Vx\times10^{- 3}\text{ }V. The value of xx is ____\_\_\_\_.

  2. Question 2

    1 Feb 2023

    57
    At what pHpH, given half cell MnO4−(0.1M)∣Mn2+(0.001M)MnO_{4}^{-}(0.1M) \mid Mn^{2 +}(0.001M) will have electrode potential of 1.282 V1.282\text{ }V ? (Nearest Integer)
     Given EMnO4∣−Mn+2o=1.54 V,2.303RTF=0.059 V\text{~Given~}E_{MnO_{4 \mid}^{-}Mn^{+ 2}}^{o}= 1.54\text{ }V,\frac{2.303RT}{F}= 0.059\text{ }V

  3. Question 3

    22 Jan 2026 S2

    46
    Consider the following electrochemical cell :
    Pt∣O2(g)(lbar)∣HCl(aq)∥M2+(aq,1.0M)∣M(s)Pt\left| O_{2}(g)(lbar) \right|HCl(aq)\| M^{2 +}(aq,1.0M) \mid M(s)
    The pH above which, oxygen gas would start to evolve at anode is ____\_\_\_\_ (nearest integer).
    [ Given :  EM2+/M0=0.994 VEO2/H2O0=1.23 V} standard reduction potential  and RTF(2.303)=0.059 V at the given condition ]\begin{bmatrix} \text{~Given :~} & \left. \ \begin{matrix} E_{M^{2 +}/M}^{0} = 0.994\text{ }V \\ E_{O_{2}/H_{2}O}^{0} = 1.23\text{ }V \end{matrix} \right\}\text{~standard reduction potential~} \\ & \text{~and~}\frac{RT}{F}(2.303) = 0.059\text{ }V\text{~at the given condition~} \end{bmatrix}