Question

use the following diagram to answer questions 54 to 56.

1. which of the following represents the overall cell reaction?

a. $ce{cr_{2}o^{2-}_{7} + h^{+} + ag -> ag^{+} + cr^{3+} + h_{2}o}$
b. $ce{cr_{2}o^{2-}_{7} + 14h^{+} + 9ag -> 9ag^{+} + cr^{3+} + 7h_{2}o}$
c. $ce{cr_{2}o^{2-}_{7} + 14h^{+} + 6ag -> 6ag^{+} + 2cr^{3+} + 7h_{2}o}$
d. $ce{cr_{2}o^{2-}_{7} + 14h^{+} + 6ag^{+} -> 6ag + 2cr^{3+} + 7h_{2}o}$

1. what happens to the ph at each electrode?

	ph at anode	ph at cathode
b.	increases	increases
c.	stays the same	decreases
d.	stays the same	increases

1. what is the cell voltage at equilibrium?

a. $-0.43$ v
b. $0.00$ v
c. $+0.43$ v
d. $+2.03$ v

Explanation:

Question 54

Step1: Identify half - reactions

• Reduction (Cathode, Pt side): In the acidic solution, \(Cr_2O_7^{2 -}\) is reduced. The reduction half - reaction of \(Cr_2O_7^{2-}\) in acidic medium is \(Cr_2O_7^{2-}+14H^{+}+6e^-

ightarrow2Cr^{3 +}+7H_2O\) (Cr goes from +6 in \(Cr_2O_7^{2-}\) to +3 in \(Cr^{3+}\), gain of 6 electrons for 2 Cr atoms, total gain of 6 \(e^-\)).

• Oxidation (Anode, Ag side): Ag is oxidized. The oxidation half - reaction is \(Ag

ightarrow Ag^{+}+e^-\) (Ag goes from 0 to +1, loss of 1 electron per Ag atom).

Step2: Balance electrons and add half - reactions

To balance the electrons, we multiply the oxidation half - reaction by 6 (since the reduction half - reaction gains 6 electrons). So the oxidation half - reaction becomes \(6Ag
ightarrow6Ag^{+}+6e^-\).

Now, add the two half - reactions:
Reduction: \(Cr_2O_7^{2-}+14H^{+}+6e^-
ightarrow2Cr^{3 +}+7H_2O\)
Oxidation: \(6Ag
ightarrow6Ag^{+}+6e^-\)
Overall reaction: \(Cr_2O_7^{2-}+14H^{+}+6Ag
ightarrow6Ag^{+}+2Cr^{3 +}+7H_2O\)

Step1: Analyze anode (Ag electrode)

At the anode, the oxidation reaction is \(Ag
ightarrow Ag^{+}+e^-\). The reaction at the anode does not involve \(H^+\) or \(OH^-\) directly. The solution in the anode compartment is \(1.0\ M\ AgNO_3\), and the reaction here is just the oxidation of Ag. So the concentration of \(H^+\) (from the salt bridge, but the anode reaction itself doesn't consume or produce \(H^+\)) remains the same. So the pH at the anode stays the same.

Step2: Analyze cathode (Pt electrode)

At the cathode, the reduction reaction is \(Cr_2O_7^{2-}+14H^{+}+6e^-
ightarrow2Cr^{3 +}+7H_2O\). In this reaction, \(H^+\) ions are consumed. As \(H^+\) concentration decreases, \(pH=-\log[H^+]\) increases (since lower \(H^+\) concentration means higher pH).

Step1: Recall cell voltage at equilibrium

For a cell, at equilibrium, the Gibbs free energy change \(\Delta G = 0\). The relationship between cell voltage (\(E_{cell}\)) and Gibbs free energy (\(\Delta G\)) is given by \(\Delta G=-nFE_{cell}\). When \(\Delta G = 0\) (at equilibrium), \(E_{cell}=0\ V\). This is because at equilibrium, there is no net flow of electrons, and the cell potential (voltage) is zero.

Answer:

C. \(\ce{Cr_{2}O^{2-}_{7} + 14H^{+} + 6Ag
ightarrow 6Ag^{+} + 2Cr^{3+} + 7H_{2}O}\)

Question 55