For the working galvanic cell Mg | Mg^2+ (1 M) || Ag+ (1 M) | Ag (E deg(Mg^2+/Mg) = -2.37 V, E deg(Ag+/Ag) = 0.80 V), which of the following statements are correct?

Electrons flow from the Mg electrode to the Ag electrode through the external wire

Anions from the salt bridge move into the Mg half cell

The reaction at the cathode is Ag+ + e- -> Ag

Electrons flow from the Mg electrode to the Ag electrode through the external wire

Anions from the salt bridge move into the Mg half cell

The mass of the Ag electrode decreases as the cell works

The reaction at the cathode is Ag+ + e- -> Ag

JEE Main levelOne or more correct optionsDifficulty 2 of 3ec035

For the working galvanic cell $Mg ∣ Mg X^{2 +} (1 M) ∣∣ Ag X^{+} (1 M) ∣ Ag$ ( $E^{\circ} (Mg X^{2 +} / Mg) = - 2.37 V$ , $E^{\circ} (Ag X^{+} / Ag) = 0.80 V$ ), which of the following statements are correct?

AElectrons flow from the Mg electrode to the Ag electrode through the external wire
BAnions from the salt bridge move into the Mg half cell
CThe mass of the Ag electrode decreases as the cell works
DThe reaction at the cathode is $Ag X^{+} + e X^{-} Ag$

Mg has the lower

E^{\circ}

, so it is the anode:

Mg Mg X^{2 +} + 2 e X^{-}

, and electrons travel through the wire to the Ag cathode where

Ag X^{+} + e X^{-} Ag

deposits silver, so the Ag electrode gains mass (C is wrong). The anode compartment accumulates

Mg X^{2 +}

, so anions from the salt bridge migrate into it to keep it neutral. A, B and D are correct.