Consider the cell Pt | H2 (1 bar) | H+ (1 M) || Cu^2+ (1 M) | Cu with E deg_cell = 0.34 V at 298 K. Which statements are correct?

The standard hydrogen electrode is the anode of this cell

If the cell were written as Cu | Cu^2+ (1 M) || H+ (1 M) | H2 (1 bar) | Pt, its EMF by convention would be -0.34 V

The reaction at the left electrode is H2 -> 2H+ + 2e-

The standard hydrogen electrode is the anode of this cell

If the cell were written as Cu | Cu^2+ (1 M) || H+ (1 M) | H2 (1 bar) | Pt, its EMF by convention would be -0.34 V

The reaction at the left electrode is H2 -> 2H+ + 2e-

If the Cu electrode were replaced by Zn | Zn^2+ (1 M) (E deg = -0.76 V), the SHE would still be the anode

JEE Advanced levelOne or more correct optionsDifficulty 3 of 3ec038

Consider the cell $Pt ∣ H X_{2} (1 bar) ∣ H X^{+} (1 M) ∣∣ Cu X^{2 +} (1 M) ∣ Cu$ with $E_{cell}^{\circ} = 0.34 V$ at 298 K. Which statements are correct?

AThe standard hydrogen electrode is the anode of this cell
BIf the cell were written as $Cu ∣ Cu X^{2 +} (1 M) ∣∣ H X^{+} (1 M) ∣ H X_{2} (1 bar) ∣ Pt$ , its EMF by convention would be $- 0.34 V$
CThe reaction at the left electrode is $H X_{2} 2 H X^{+} + 2 e X^{-}$
DIf the Cu electrode were replaced by $Zn ∣ Zn X^{2 +} (1 M)$ ( $E^{\circ} = - 0.76 V$ ), the SHE would still be the anode

E_{cell}^{\circ} = E_{right}^{\circ} - E_{SHE}^{\circ} = 0.34 - 0 = + 0.34 V

, so the cell is spontaneous as written: the SHE (left) is the anode and hydrogen is oxidised,

H X_{2} 2 H X^{+} + 2 e X^{-}

. Reversing the notation reverses the sign, giving

- 0.34 V

. With Zn,

E^{\circ} (Zn X^{2 +} / Zn) = - 0.76 V < 0

, so Zn is oxidised and the SHE becomes the cathode, so D is wrong.