91Ó°ÊÓ

The half-reactions for iron and chromium ions are as follows: Iron(II) ions to form iron metal: Fe^2+ (aq) → Fe(s) Chromium metal to form chromium(III) ions: Cr(s) → Cr^3+ (aq)

In order to balance the half-reactions, we must ensure that both the atoms and charges are balanced. For iron half-reaction: Fe^2+ (aq) → Fe(s) It is already balanced. For chromium half-reaction: Initial: Cr(s) → Cr^3+ (aq) Balance the charge by adding 3 electrons to the right side: Cr(s) → Cr^3+ (aq) + 3e^-

To get the overall reaction, we need to make the electron transfer equal in both half-reactions. Since we have 3 electrons in the chromium half-reaction, we must multiply the iron half-reaction by 3 to balance the electrons: 3(Fe^2+ (aq) → Fe(s)) Resulting in: 3Fe^2+ (aq) → 3Fe(s) + 6e^- Now we can add the balanced half-reactions to get the overall reaction: 3Fe^2+ (aq) + Cr(s) → 3Fe(s) + Cr^3+ (aq)

To calculate the E° cell or the voltage, we can use the formula: E° cell = E° cathode - E° anode We know that Fe^2+ (aq) → Fe(s) is the reduction half-reaction (formation of iron metal) and Cr(s) → Cr^3+ (aq) is the oxidation half-reaction (changing chromium metal to chromium(III)). Look up their standard reduction potentials in a table: Fe^2+ (aq) + 2e^- → Fe(s), E°= -0.44 V Cr^3+ (aq) + 3e^- → Cr(s), E°= -0.74 V Substitute into the formula: E° cell = E°(Fe^2+/Fe) - E°(Cr^3+/Cr) = (-0.44) - (-0.74) = 0.30 V

To sketch the galvanic cell, follow these guidelines: - Label the anode (oxidation) and cathode (reduction) sides of the cell - Show the electron flow from the anode to the cathode - Include the salt bridge and anions and cations Here is the sketch description of the galvanic cell: Left side (Anode): Chromium metal electrode where oxidation of chromium metal occurs: Cr(s) → Cr^3+ (aq) + 3e^- Right side (Cathode): Iron metal electrode where reduction of iron(II) ions occurs: 3Fe^2+ (aq) + 6e^- → 3Fe(s) Salt bridge: Connect the two half-cells, allowing ion flow between them to maintain charge neutrality Electron flow: Electrons flow from chromium metal (anode) to iron metal (cathode) Anions and cations: Flow towards appropriate sides (anode or cathode) via the salt bridge; Cr^3+ ions produced at the anode side and Fe^2+ ions reduced at the cathode side.