Question

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3 Question 44 (1 point) What is the standard Gibbs free energy change for the reaction below? 2 Au + 3 Mn2+ - 2 Au3+ + 3 Mn % reaction Ered (V) AU3+ + 30 - Au +1.50 Mn2+ + 2e -- Mn -1.18 +185 kJ 0-517 kJ 0-185 kJ -1.55 x 103 kJ +517 kJ none of these +1.55 x 103 kJ

Answer 1

Question 44

Answer:

The standard gibbs free energy change ($\small \Delta$G⁰) = +1.55 $\small \times$ 10³ KJ.

So, Option 7. +1.55 $\small \times$ 10³ KJ is the correct answer.

Calculation:

Anode half cell reaction is:

Au(s) - 3e^-$\small \rightarrow$ Au³⁺(aq).........(1) E⁰_anode = 1.50 V

cathode half cell reaction is:

Mn²⁺(aq) + 2e^-$\small \rightarrow$ Mn(s)....(2) E⁰_cathode = -1.18 V

From (1)*2 and (2)*3 we get,

2Au(s) - 6e^-$\small \rightarrow$ 2Au³⁺(aq).........(3) E⁰_anode = 1.50 V

3Mn²⁺(aq) + 6e^-$\small \rightarrow$ 3Mn(s)....(4) E⁰_cathode = -1.18 V

By adding equation (3) and (4),

2Au(s) + 3Mn²⁺(aq) $\small \rightarrow$ 2Au³⁺(aq) + 3Mn(s)

Now, E⁰ = E⁰_cathode - E⁰_anode

Or, E⁰ = (- 1.18 V) - (1.50 V) = (-1.18 -1.50) V = -2.68 V

Now, $\small \Delta$ G⁰ = -nFE⁰

Where,  $\small \Delta$G⁰ is standard gibbs free energy change; n is number e^-s involve in the reaction = 6; F is faraday constant = 96.5 KJ/V.

Now, putting the the value in the above equation we get,

$\small \Delta$G⁰ = -nFE⁰

Or, $\small \Delta$ G⁰ = -(6) $\small \times$ (96.5 KJ/V) $\small \times$ (-2.68 V)

Or, $\small \Delta$ G⁰ = +1.55 $\small \times$ 10³ KJ (answer).

Question 43

Answer:

The TRUE statements are:

Option 1. Oxidation occurs at the anode.

Option 2. In a voltaic cell electron flow is from the anode to cathode.

Option 3. The oxidation half-reaction is always on the left side of the cell.

Explanation:

In electrochemistry, the anode is where oxidation occurs and electrons always flow from the anode to the cathode. The anode (oxidation half-reaction) is always placed on the left side, and the cathode is placed on the right side.

Other two options are not correct as,

The ions transfer from one electrode to another by a salt bridge not to move electrons.

The inert electrode can act as an electrocatalyst not reactant or product.

Question 42

Answer:

3 electrons are added on reactant side.

So, Option 6. 3 electrons on reactant side is the correct answer.

Calculation:

The given equation is,

NO₃^-$\small \rightarrow$ NO

The balanced equation in acidic solution,

NO₃^- + 4H⁺ + 3e^-$\small \rightarrow$ NO + 2H₂O

So 3 electrons are added on reactant side.

Question 41

Answer:

Cu is the cathode in the voltaic cell.

So, Option 1. Cu is the correct answer.

Explanation:

Generally we use Au for the anode because it needs a solid conducting metal for the electrode. So, Cu is a cathode which deposited as a metal solid on the electrode surface.