Analyzing a dichromate cobalt galvanic cell notation

Key takeaway: This cell combines a cobalt metal half-cell with an acidic dichromate half-cell, so the key is to compare standard reduction tendencies before writing the half-reactions.

Identify the half-cells and redox roles

The cell notation is

$|\text{Co}(s)|\text{Co}^{2+}(aq)||\text{Cr}_2\text{O}_7^{2-}(aq),\text{H}^+(aq)|\text{C}(s)|$

This tells us that cobalt metal is in contact with $\text{Co}^{2+}(aq)$ on the left, while acidic dichromate is reduced at an inert carbon electrode on the right. The most important step is to compare the standard reduction potentials.

For cobalt:

$\text{Co}^{2+}(aq)+2e^-\rightarrow \text{Co}(s)$

For acidic dichromate:

$\text{Cr}_2\text{O}_7^{2-}(aq)+14\text{H}^+(aq)+6e^-\rightarrow 2\text{Cr}^{3+}(aq)+7\text{H}_2\text{O}(l)$

Because dichromate has the much more positive reduction potential, it is reduced at the cathode. Cobalt is therefore oxidized at the anode.

Write the cathode, anode, and overall reaction

Anode (oxidation):

$\text{Co}(s)\rightarrow \text{Co}^{2+}(aq)+2e^-$

Multiply by 3 to balance electrons:

$3\text{Co}(s)\rightarrow 3\text{Co}^{2+}(aq)+6e^-$

Cathode (reduction):

$\text{Cr}_2\text{O}_7^{2-}(aq)+14\text{H}^+(aq)+6e^-\rightarrow 2\text{Cr}^{3+}(aq)+7\text{H}_2\text{O}(l)$

Add the half-equations:

$3\text{Co}(s)+\text{Cr}_2\text{O}_7^{2-}(aq)+14\text{H}^+(aq)\rightarrow 3\text{Co}^{2+}(aq)+2\text{Cr}^{3+}(aq)+7\text{H}_2\text{O}(l)$

So the net cell reaction is the balanced equation above.

Predict the strongest oxidizing and reducing agents, and the cell potential

The strongest oxidizing agent is the species that gets reduced: $\text{Cr}_2\text{O}_7^{2-}$ in acidic solution. The strongest reducing agent is the species that gets oxidized: $\text{Co}(s)$.

Using standard reduction potentials:

• $E^\circ(\text{Cr}_2\text{O}_7^{2-}/\text{Cr}^{3+}) = +1.33\text{ V}$
• $E^\circ(\text{Co}^{2+}/\text{Co}) = -0.28\text{ V}$

Then

$E^\circ_{\text{cell}}=E^\circ_{\text{cathode}}-E^\circ_{\text{anode}}=1.33-(-0.28)=1.61\text{ V}$

So the predicted standard cell potential is $+1.61$ V.

Diagram details, electrode labels, and ion flow

A correct cell diagram should show:

• Anode (left): Co(s) strip dipped in $\text{Co}^{2+}(aq)$
• Cathode (right): inert C(s) electrode dipped in acidic solution containing $\text{Cr}_2\text{O}_7^{2-}(aq)$ and $\text{H}^+(aq)$
• Electron flow: from Co anode to carbon cathode through the external wire
• Ion flow through the salt bridge: anions migrate toward the anode compartment to balance the buildup of $\text{Co}^{2+}$, while cations migrate toward the cathode compartment to replace positive charge being consumed there

A clear labeled sketch would show oxidation at the cobalt electrode and reduction at the carbon electrode. Since carbon is inert, it does not react; it only provides a surface for electron transfer.

For presentation, you can label the cell as:

$\text{Co}(s)|\text{Co}^{2+}(aq)||\text{Cr}_2\text{O}_7^{2-}(aq),\text{H}^+(aq)|\text{C}(s)$

That notation matches the redox direction, electrode identity, and phase boundaries.

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Pitfalls the pros know 👇 A common mistake is to treat the carbon electrode as a reacting species. In this cell, carbon is only an inert conductor, so it is not the oxidizing or reducing agent. Another frequent error is writing the dichromate half-reaction without acid; in acidic solution, $\text{H}^+$ and water must appear to balance both hydrogen and oxygen. Students also sometimes reverse anode and cathode because they remember the left side of notation as the anode without checking reduction potentials. Always confirm by comparing $E^\circ$ values, then balance electrons before adding half-reactions. Finally, do not forget that the net reaction must cancel electrons and must not leave spectator species from the half-cell notation unless they are actually used in balancing.

What if the problem changes? If the same cell were written in basic solution instead of acidic solution, the dichromate half-reaction would change because $\text{H}^+$ would no longer be available. The variant would need $\text{OH}^-$ and water to balance the reduction of $\text{Cr}_2\text{O}_7^{2-}$ to $\text{Cr}^{3+}$. For example, a related prompt might be: 'For the cell $|\text{Co}(s)|\text{Co}^{2+}(aq)||\text{Cr}_2\text{O}_7^{2-}(aq),\text{OH}^-(aq)|\text{C}(s)|$, identify the half-reactions and predict $E^\circ$.' In that case, the cobalt half-reaction stays the same, but the cathode equation must be rewritten for alkaline conditions. The identity of the oxidizing agent is still dichromate, and cobalt is still the reducing agent, but the balancing method changes because the medium changes. That is why the solution strategy must always start by checking whether the half-cell is acidic, basic, or neutral.

Tags: standard reduction potential, half-cell notation, salt bridge