Question

question 14 of 30 what is the cell potential of an electrochemical cell that has the half - reactions shown below? ag⁺+e⁻→ag fe→fe³⁺ + 3e⁻ click for a reduction potential chart a. 0.44 v b. -0.44 v c. -1.24 v d. 1.24 v

Explanation:

Step1: Identify oxidation and reduction half - reactions

The silver half - reaction $Ag^+ + e^-\to Ag$ is a reduction (gain of electrons), and the iron half - reaction $Fe\to Fe^{3+}+3e^-$ is an oxidation (loss of electrons).

Step2: Find standard reduction potentials

From the reduction potential chart, $E^0_{Ag^+/Ag}= + 0.80\ V$ and $E^0_{Fe^{3+}/Fe}=- 0.04\ V$.

Step3: Calculate the cell potential

The formula for the cell potential $E^0_{cell}=E^0_{cathode}-E^0_{anode}$. The cathode is where reduction occurs (silver half - reaction) and the anode is where oxidation occurs (iron half - reaction). So $E^0_{cell}=E^0_{Ag^+/Ag}-E^0_{Fe^{3+}/Fe}=0.80-(- 0.44)=1.24\ V$.

Answer:

D. 1.24 V