Question

Consider the following reaction undwhat is the standard cell potential for a galvanic cell that consists of ag /ag and fe2 /fe half-cells? the reactions involved in the galvanic cell, both written as reductions, are ag (aq) e−⟶ag(s)e∘ag /ag= 0.80 vfe2 (aq) 2e−⟶fe(s)e∘fe2 /fe=−0.44 ver basic conditions: mno−4(aq) al(s)⟶[al(oh)4]−(aq) mno2(s) how many water molecules will appear in the balanced equation?

Answer 1

The standard cell potential for a galvanic cell consisting of Ag/Ag+ and Fe2+/Fe half-cells is calculated by subtracting the reduction potential of the anode from the reduction potential of the cathode, resulting in a value of 1.24 V.

Step-by-step explanation:

To calculate the standard cell potential for the given galvanic cell consisting of Ag/Ag+ and Fe2+/Fe half-cells, we need to use the given reduction potentials for these half-reactions:

Ag+ + e− ⇒ Ag(s) E°Ag/Ag+ = 0.80 V
Fe2+ (aq) + 2e− ⇒ Fe(s) E°Fe2+/Fe = -0.44 V

The standard cell potential (E°cell) is calculated by subtracting the potential for the oxidation half-reaction (anode) from the potential for the reduction half-reaction (cathode). However, since all potentials are given in the form of reduction potentials, we take the negative of the anode's reduction potential to represent its oxidation potential. In this case, the Fe electrode is the anode and the Ag electrode is the cathode since the more positive potential is always the cathode.

The overall standard cell potential is then:

E°cell = E°cathode - E°anode
E°cell = 0.80 V - (-0.44 V)
E°cell = 0.80 V + 0.44 V
E°cell = 1.24 V

Therefore, the standard cell potential for the galvanic cell is 1.24 V.