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Standard potential of a third couple from two given potentials (adding via Gibbs energy)

This type of question, asked 6 times: 28 Jun 2022, 10 Apr 2023, 11 Apr 2023, 23 Jan 2025, 24 Jan 2025, 8 Apr 2026 S2. All 6 questions below, oldest first.

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  1. Question 1

    28 Jun 2022

    144
    For the given reactions
    Sn2++2e−→SnSn^{2 +}+ 2e^{-}\rightarrow Sn
    Sn4++4e−→SnSn^{4 +}+ 4e^{-}\rightarrow Sn
    The electrode potentials are; ESn2+/Sno=−0.140 VE_{Sn^{2 +}/Sn}^{o}= - 0.140\text{ }V and ESn4+/Sno=0.010 VE_{Sn^{4 +}/Sn}^{o}= 0.010\text{ }V. The magnitude of standard electrode potential for Sn4+/Sn2+Sn^{4 +}/Sn^{2 +} i.e. ESn4+/Sn2+oE_{Sn^{4 +}/Sn^{2 +}}^{o} is ____×10−2 V\_\_\_\_\times10^{- 2}\text{ }V. (Nearest integer)

  2. Question 2

    10 Apr 2023

    56
    FeO42−→+2.2 VFe3+→+0.70 VFe2+→−0.45 VFe0FeO_{4}^{2 -}\overset{+ 2.2\text{ }V}{\rightarrow}Fe^{3 +}\overset{+ 0.70\text{ }V}{\rightarrow}Fe^{2 +}\overset{- 0.45\text{ }V}{\rightarrow}Fe^{0} EFeO42−/Fe2+θE_{FeO_{4}^{2 -}/Fe^{2 +}}^{\theta} is x×10−3 Vx \times10^{- 3}\text{ }V. The value of xx is__

  3. Question 3

    11 Apr 2023

    60
    In an electrochemical reaction of lead, at standard temperature, if E(Pb2+/Pb)0=mE_{\left( Pb^{2 +}/Pb \right)}^{0}= m Volt and E(Pb4+/Pb)0=nE_{\left( Pb^{4 +}/Pb \right)}^{0}= n Volt , then the value of E(Pb2+/Pb4+)0E_{\left( Pb^{2 +}/Pb^{4 +} \right)}^{0} is given by m−xnm - xn. The value of xx is___ (Nearest integer)

  4. Question 4

    23 Jan 2025

    30
    FeO42−→+2.0VFe3+→0.8VFe2+→−0.5VFe0FeO_{4}^{2 -}\overset{+ 2.0V}{\rightarrow}Fe^{3 +}\overset{0.8V}{\rightarrow}Fe^{2 +}\overset{- 0.5V}{\rightarrow}Fe^{0} In the above diagram, the standard electrode potentials are given in volts (over the arrow). The value of EFeO42−/Fe2+⊖E_{FeO_{4}^{2 -}/Fe^{2 +}}^{\ominus} is

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  5. Question 5

    24 Jan 2025

    26
    For the given cell
    Fe2+(aq)+Ag+(aq)→Fe3+(aq)+Ag( s)Fe^{2 +}(aq) +Ag^{+}(aq) \rightarrow Fe^{3 +}(aq) + Ag(\text{ }s)
    The standard cell potential of the above reaction is Given :
    Ag++e−→Ag   E0=xVFe2++2e−→Fe   E0=yVFe3++3e−→Fe    E0=zV\begin{matrix} Ag^{+}+e^{-}\rightarrow Ag & \ \ \ E^{0}= xV \\ Fe^{2 +}+ 2e^{-}\rightarrow Fe & \ \ \ E^{0}= yV \\ Fe^{3 +}+ 3e^{-}\rightarrow Fe & \ \ \ \ E^{0}= zV \end{matrix}

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  6. Question 6

    8 Apr 2026 S2

    31
    Given at 298 K:EFe2+/Fe⊖=X298\text{ }K:E_{Fe^{2 +}/Fe}^{\ominus}= X Volt
    EFe3+/Fe⊖=Y Volt E_{Fe^{3 +}/Fe}^{\ominus}= Y\text{~Volt~}
    The EFe3+/Fe2+⊖E_{Fe^{3 +}/Fe^{2 +}}^{\ominus} in Volt at 298 K is given by :

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