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MISSED THIS? Read Section 20.6 (Pages 917-922). An electrochemical cell is based on the following two half-reactions: oxidation: Sn (s) $\rightarrow$ Sn$^{2+}$ (aq, 1.70 M)+2e$^-$ reduction: Ag$^+$ (aq, 0.170 M)+e$^-$ $\rightarrow$ Ag(s) Part A Compute the cell potential at 25 $^\circ$C. Express the cell potential to two decimal places. $E_{cell}$ = 0.94 V Incorrect; Try Again; One attempt remaining You have calculated the standard cell potential. Use the Nernst equation to calculate the cell potential under the conditions given above: $E_{cell} = E^\circ_{cell} - \frac{0.0592}{n} \log Q$ where $E_{cell}$ is the cell potential in volts, $E^\circ_{cell}$ is the standard cell potential in volts, $n$ is the number of moles of electrons transferred in the redox reaction, and $Q$ is the reaction quotient.

          MISSED THIS? Read Section
20.6 (Pages 917-922).
An electrochemical cell is based on the following
two half-reactions:
oxidation: Sn (s) $\rightarrow$ Sn$^{2+}$ (aq, 1.70 M)+2e$^-$
reduction: Ag$^+$ (aq, 0.170 M)+e$^-$ $\rightarrow$ Ag(s)
Part A
Compute the cell potential at 25 $^\circ$C.
Express the cell potential to two decimal places.
$E_{cell}$ = 0.94
V
Incorrect; Try Again; One attempt remaining
You have calculated the standard cell potential. Use the Nernst equation to calculate the cell potential under the
conditions given above:
$E_{cell} = E^\circ_{cell} - \frac{0.0592}{n} \log Q$
where $E_{cell}$ is the cell potential in volts, $E^\circ_{cell}$ is the standard cell potential in volts, $n$ is the number of moles of
electrons transferred in the redox reaction, and $Q$ is the reaction quotient.

Show more…

MISSED THIS? Read Section
20.6 (Pages 917-922).
An electrochemical cell is based on the following
two half-reactions:
oxidation: Sn (s) → Sn^2+ (aq, 1.70 M)+2e^-
reduction: Ag^+ (aq, 0.170 M)+e^- → Ag(s)
Part A
Compute the cell potential at 25 ^∘C.
Express the cell potential to two decimal places.
Ecell = 0.94
V
Incorrect; Try Again; One attempt remaining
You have calculated the standard cell potential. Use the Nernst equation to calculate the cell potential under the
conditions given above:
Ecell = E^∘cell - (0.0592)/(n)log Q
where Ecell is the cell potential in volts, E^∘cell is the standard cell potential in volts, n is the number of moles of
electrons transferred in the redox reaction, and Q is the reaction quotient.

Added by Andrea V.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert David Collins

04/16/2023

Step 1

E°cell = E°reduction - E°oxidation E°cell = 0.80 V - (-0.14 V) E°cell = 0.94 V Show more…

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13 of 16 Review I Constants I Periodic Table MISSED THIS? Read Section 20.6 (Pages 917 - 922). An electrochemical cell is based on the following two half-reactions: oxidation: Sn(s) -> Sn^(2+)(aq, 1.70M) + 2e^(-) reduction: Ag^(+)(aq, 0.170M) + e^(-) -> Ag(s) Part A Compute the cell potential at 25°C. Express the cell potential to two decimal places. E_(cell) = ___, V Previous Answers Request Answer Incorrect; Try Again; One attempt remaining You have calculated the standard cell potential. Use the Nernst equation to calculate the cell potential under the conditions given above: E_(cell) = E_(cell)° - (0.0592)/(n)logQ where E_(cell) is the cell potential in volts, E_(cell)° is the standard cell potential in volts, n is the number of moles of electrons transferred in the redox reaction, and Q is the reaction quotient. Provide Feedback 13 of 16 Review I Constants I Periodic Table Part A MISSED THIS? Read Section 20.6 (Pages 917-922). An electrochemical cell is based on the following two half-reactions: oxidation: Sn(s) -> Sn^(2+)(aq, 1.70 M) + 2e reduction: Ag^(+)(aq, 0.170 M) + e -> Ag(s) Compute the cell potential at 25°C. Express the cell potential to two decimal places. E_(cell) = 0.94 Submit Previous Answers Request Answer Incorrect; Try Again; One attempt remaining You have calculated the standard cell potential. Use the Nernst equation to calculate the cell potential under the conditions given above: E_(cell) = E_(cell) - 0.0210g electrons transferred in the redox reaction, and Q is the reaction quotient. Provide Feedback Next

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