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What is the standard cell notation for an electrolytic cell with aluminum and gold electrodes?A.Au+(aq) | Au(s) || Al(s) | Al3+(aq)B.Al3+(aq) | Al(s) || Au(s) | Au+(aq)C.Al(s) | Al3+(aq) || Au+(aq) | Au(s)D.Au(s) | Au+(aq) || Al3+(aq) | Al(s)SUBMITarrow_backPREVIOUS

Question

What is the standard cell notation for an electrolytic cell with aluminum and gold electrodes?A.Au+(aq) | Au(s) || Al(s) | Al3+(aq)B.Al3+(aq) | Al(s) || Au(s) | Au+(aq)C.Al(s) | Al3+(aq) || Au+(aq) | Au(s)D.Au(s) | Au+(aq) || Al3+(aq) | Al(s)SUBMITarrow_backPREVIOUS

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Solution

The standard cell notation for an electrolytic cell is written as: Anode | Anode Ion || Cathode Ion | Cathode.

In an electrolytic cell, oxidation occurs at the anode and reduction occurs at the cathode.

Aluminum (Al) will be oxidized to Al3+ because it is more reactive than gold (Au). This means Al will be the anode.

Gold (Au) will be reduced to Au+ because it is less reactive, so Au will be the cathode.

Therefore, the correct notation would be: Al(s) | Al3+(aq) || Au+(aq) | Au(s).

So, the answer is C. Al(s) | Al3+(aq) || Au+(aq) | Au(s).

This problem has been solved

Similar Questions

What are the half-reactions for a galvanic cell with aluminum and gold electrodes?A.Al(s) Al3+(aq) + 3e- and Au+(aq) + e- Au(s)B.Al(s) Al3+(aq) + 3e- and Au(s) Au+(aq) + e-C.Al3+(aq) + 3e- Al(s) and Au+(aq) + e- Au(s)D.Al3+(aq) + 3e- Al(s) and Au(s) Au+(aq) + e-

. Choose the correct line notation for the following galvanic cell reaction:2Al(s) + 3Cu+2(aq) → Al+3(aq) + 3Cu(s)Line Notation: Cu2+(aq) | Cu(s) || Al(s) | Al3+(aq)Al3+(aq) | Al(s) || Cu(s) | Cu2+(aq)Al(s) | Al3+(aq) || Cu2+(aq) | Cu(s)

To write the shorthand notation for the given galvanic cell, we need to follow these steps: 1. **Identify the anode and cathode reactions:** - **Anode (oxidation):** Lead (Pb) is oxidized to lead ions (Pb²⁺). \[ \text{Pb(s)} \rightarrow \text{Pb}^{2+}(\text{aq}) + 2\text{e}^- \] - **Cathode (reduction):** Ferric ions (Fe³⁺) are reduced to ferrous ions (Fe²⁺). \[ \text{Fe}^{3+}(\text{aq}) + \text{e}^- \rightarrow \text{Fe}^{2+}(\text{aq}) \] 2. **Write the cell notation:** - The anode (oxidation) half-cell is written on the left. - The cathode (reduction) half-cell is written on the right. - A single vertical line (|) separates different phases within the same half-cell. - A double vertical line (||) represents the salt bridge between the two half-cells. Given the reactions: - Anode: \(\text{Pb(s)} \rightarrow \text{Pb}^{2+}(\text{aq}) + 2\text{e}^-\) - Cathode: \(\text{Fe}^{3+}(\text{aq}) + \text{e}^- \rightarrow \text{Fe}^{2+}(\text{aq})\) The shorthand notation for the galvanic cell is: \[ \text{Pb(s)} | \text{Pb}^{2+}(\text{aq}) || \text{Fe}^{3+}(\text{aq}), \text{Fe}^{2+}(\text{aq}) | \text{Pt(s)} \] Here: - \(\text{Pb(s)}\) is the solid lead electrode. - \(\text{Pb}^{2+}(\text{aq})\) is the lead ion in solution. - \(\text{Fe}^{3+}(\text{aq})\) and \(\text{Fe}^{2+}(\text{aq})\) are the ferric and ferrous ions in solution. - \(\text{Pt(s)}\) is the platinum electrode used as an inert conductor for the cathode reaction

i) Write the electronic configuration of an Al 3+ ion

In a zinc-copper cell, Zn | Zn(1M) || Cu(1M) | Cu, which electrode is negative?Group of answer choicesCu(s)Zn(s)Zn(aq)Cu(aq)

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