Will the concentration of Ni2+ in the Ni2+ - Ni half-cell increase or decrease as the cell operates?

Verified step by step guidance

Step 1: Understand the context of a galvanic cell. In a galvanic cell, a spontaneous redox reaction generates electrical energy. The cell consists of two half-cells, each containing an electrode and an electrolyte solution.

Step 2: Identify the role of the Ni2+ - Ni half-cell. In this half-cell, nickel metal (Ni) is in equilibrium with nickel ions (Ni^{2+}) in solution. The reaction can be written as: Ni(s) \rightleftharpoons Ni^{2+}(aq) + 2e^{-}.

Step 3: Determine the direction of electron flow. In a galvanic cell, electrons flow from the anode to the cathode. The anode is where oxidation occurs, and the cathode is where reduction occurs.

Step 4: Analyze the oxidation process. If the Ni2+ - Ni half-cell is the anode, oxidation occurs here, meaning Ni(s) loses electrons to form Ni^{2+}(aq). This increases the concentration of Ni^{2+} ions in the solution.

Step 5: Conclude the effect on Ni2+ concentration. As the cell operates and oxidation continues at the anode, the concentration of Ni^{2+} ions in the Ni2+ - Ni half-cell will increase.

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Electrochemical Cells

Electrochemical cells consist of two half-cells where oxidation and reduction reactions occur. In a half-cell, the concentration of ions can change as the cell operates, affecting the overall cell potential. Understanding how these reactions influence ion concentration is crucial for predicting changes in the cell's behavior.

Nernst Equation

The Nernst equation relates the cell potential to the concentrations of the reactants and products in an electrochemical reaction. It shows that as the concentration of Ni2+ changes, the cell potential will also change, allowing us to predict whether the concentration will increase or decrease during operation.

Le Chatelier's Principle

Le Chatelier's Principle states that if a system at equilibrium is disturbed, the system will adjust to counteract the disturbance and restore a new equilibrium. In the context of the Ni2+ - Ni half-cell, if the concentration of Ni2+ decreases due to the reaction, the system will shift to produce more Ni2+, affecting the overall concentration as the cell operates.

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Related Practice

Textbook Question

Predict whether the following reactions will be spontaneous in acidic solution under standard conditions: (a) oxidation of Sn to Sn²⁺ by I₂ (to form I^-), (b) reduction of Ni²⁺ to Ni by I^- (to form I₂), (c) reduction of Ce⁴⁺ to Ce³⁺ by H₂O₂, (d) reduction of Cu²⁺ to Cu by Sn²⁺ (to form Sn⁴⁺).

Textbook Question

Gold exists in two common positive oxidation states, +1 and +3. The standard reduction potentials for these oxidation states are Au+1aq2 + e- ¡ Au1s2 Ered ° = +1.69 V Au3+1aq2 + 3 e- ¡ Au1s2 Ered ° = +1.50 V (c) Miners obtain gold by soaking gold-containing ores in an aqueous solution of sodium cyanide. A very soluble complex ion of gold forms in the aqueous solution because of the redox reaction 4 Au1s2 + 8 NaCN1aq2 + 2 H2O1l2 + O21g2 ¡ 4 Na3Au1CN2241aq2 + 4 NaOH1aq2 What is being oxidized, and what is being reduced in this reaction?

Textbook Question

A voltaic cell is constructed that uses the following half-cell reactions:

Cu⁺(aq) + e^- → Cu(s)

I₂(s) + 2 e^- → 2 I^-(aq)

The cell is operated at 298 K with [Cu⁺] = 0.25 M and [I^-] = 0.035 M.

(a) Determine E for the cell at these concentrations.

Textbook Question

A voltaic cell is constructed that uses the following half-cell reactions:

Cu⁺(aq) + e^- → Cu(s)

I₂(s) + 2 e^- → 2 I^-(aq)

The cell is operated at 298 K with [Cu⁺] = 0.25 M and [I^-] = 0.035 M.

(b) Which electrode is the anode of the cell?