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Spontaneous direction and standard EMF of a cell from two half-cell potentials

This type of question, asked 3 times: 25 Jun 2022, 5 Apr 2024, 2 Apr 2026 S1. All 3 questions below, oldest first.

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  1. Question 1

    25 Jun 2022

    57
    In a cell, the following reactions take place
    Fe2+→Fe3++e−EFe3+/Fe2+o=0.77 V2I−→I2+2e−EI2/I−o=0.54 V\begin{matrix} Fe^{2 +}\rightarrow Fe^{3 +}+e^{-} & E_{Fe^{3 +}/Fe^{2 +}}^{o}= 0.77\text{ }V \\ 2I^{-}\rightarrow I_{2}+ 2e^{-} & E_{I_{2}/I^{-}}^{o}= 0.54\text{ }V \end{matrix}
    The standard electrode potential for the spontaneous reaction in the cell is x×10−2 Vx \times10^{- 2}\text{ }V at 298 K298\text{ }K. The value of xx is (Nearest Integer)

  2. Question 2

    5 Apr 2024

    134
    For the electro chemical cell
    M∣M2+∣X∣X2−M\left| M^{2 +} \right|X \mid X^{2 -}
    If E(M2+/M)0=0.46 VE_{\left( M^{2 +}/M \right)}^{0}= 0.46\text{ }V and E(X/X2−)0=0.34 VE_{\left( X/X^{2 -} \right)}^{0}= 0.34\text{ }V Which of the following is correct?

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  3. Question 3

    2 Apr 2026 S1

    32
    An electrochemical cell is constructed using half cells in the direction of spontaneous change: Fe(OH)2( s)+2e−→Fe(s)+2OH−(aq) E0=−0.88 VFe(OH)_{2}(\text{ }s) + 2e^{-}\rightarrow Fe(s) + 2OH^{-}(aq)\ E^{0}= - 0.88\text{ }V and AgBr(s)+e−→Ag(s)+Br−(aq) E0=+0.07 VAgBr(s) +e^{-}\rightarrow Ag(s) +Br^{-}(aq)\ E^{0}= + 0.07\text{ }V Which of the following option is correct?

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