Question

Consider the following general voltaic cell and a cell notation, and answer all three parts of this question.

Mg(s) | Mg2+(aq) || Cl2(g) | Cl−(aq) | C(s)


Part 1: Based on the cell notation, the substance of Electrode A is [ Select ] ["Mg(s)", "C(s)", ""] , and the substance of Electrode B is [ Select ] ["C(s)", "Mg(s)"] .

Part 2: What is the balanced redox equation for the voltaic cell?

A) Mg2+(aq) + 2 Cl−(aq) \rightarrow → Mg(s) + Cl2(g)

B) Mg2+(aq) + Cl−(aq) \rightarrow → Mg(s) + Cl2(g)

C) Mg(s) + Cl2(g) \rightarrow → Mg2+(aq) + 2 Cl−(aq)

D) Mg(s) + Cl2(g) \rightarrow → Mg2+(aq) + Cl−(aq)

E) None of the above is correct, because C(s) doesn't appear in any of them.


Part 3:

- If the salt bridge contains NaNO3, Na+ ions flow to the [ Select ] ["left", "right"] , and NO3− ions flow to the [ Select ] ["left", "right"] .

- Electrons flow from Electrode [ Select ] ["A", "B"] to Electrode [ Select ] ["B", "A"] .

Answer 1

Answer:

Part -1

Electrode A - Mg(s)

Electrode B - $Cl_{2}$

Part-2

Option -C

Part - 3

Electrons flow from electrode -A to electrode -B.

Explanation:

Part-1

Electrode A is $Mg(s)|Mg^{+2}(aq)$

Electrode B is $Cl_{2}(g)|Cl^{-}(aq)|C(s)$

Part-2

From the electrode, magnesium looses two electrons and chlorine gains two electrons. Therefore, balanced redox equation for the voltaic cell is as follows.

$Mg(s)+Cl_{2}(g)\rightarrow Mg^{2+}(aq)+2Cl^{-}(aq)$

Part-3

$Na^{+}$ ions flow from the left to right.

$NO_{3}^{-}$ ions flow from the right to left.

Therefore, Electrons flow from electrode -A to electrode -B.