Question

suppose a student accidentally used naoh, instead of nacl for the salt bridge. what would be the outcome?
standard reduction potentials
half - reaction | e° (v)
mg²⁺ + 2e⁻ → mg | - 2.37
fe²⁺ + 2e⁻ → fe | - 0.44
the voltage would double.
bubbles would form in the cathode half - cell.
a precipitate would form in the anode half - cell.
there would be no change in voltage.

Explanation:

1. First, determine the anode and cathode: The half - reaction with the more negative standard reduction potential (E°) is the oxidation (anode) reaction. For \(Mg^{2 +}+2e^-

ightarrow Mg\), \(E°=- 2.37\ V\) and for \(Fe^{2+}+2e^-
ightarrow Fe\), \(E° = - 0.44\ V\). Since \(Mg\) has a more negative \(E°\), \(Mg\) is oxidized (anode: \(Mg
ightarrow Mg^{2+}+2e^-\)) and \(Fe^{2+}\) is reduced (cathode: \(Fe^{2+}+2e^-
ightarrow Fe\)).

1. Analyze the effect of \(NaOH\) in the salt bridge: In the anode half - cell (where \(Mg/Mg^{2+}\) is), \(Mg^{2+}\) ions are present. When \(NaOH\) is used in the salt bridge, \(OH^-\) ions will migrate into the anode half - cell. The reaction \(Mg^{2+}(aq)+2OH^-(aq)

ightarrow Mg(OH)_2(s)\) occurs. \(Mg(OH)_2\) is a precipitate.

1. Analyze the other options:

• Option "The voltage would double": The cell voltage is determined by the difference in standard reduction potentials of the two half - reactions (\(E_{cell}=E_{cathode}-E_{anode}\)). Using \(NaOH\) in the salt bridge does not double the difference in \(E°\) values, so this is incorrect.
• Option "Bubbles would form in the cathode half - cell": In the cathode half - cell, the reaction is \(Fe^{2+}+2e^-

ightarrow Fe\). There is no reaction that would produce gas (bubbles) here with \(NaOH\) in the salt bridge, so this is incorrect.

• Option "There would be no change in voltage": The formation of a precipitate in the anode half - cell will change the concentration of \(Mg^{2+}\) ions, which will affect the cell potential (by the Nernst equation \(E = E^0-\frac{RT}{nF}\ln Q\), where \(Q\) is the reaction quotient). Also, the physical change of precipitate formation disrupts the cell operation, so there will be a change, and this option is incorrect.

Answer:

A precipitate would form in the anode half - cell.