When [Pb2+] = 1.40 M, the observed cell potential at 298 K for an electrochemical cell with the reaction shown below is 1.150 V. What is the Mn2+ concentration in this cell? Pb2*(aq) + Mn(s)– Pb(s) + Mn²+(aq) [Mn2+] = mol/L

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When [Pb2+] = 1.40 M, the observed cell potential at 298 K for an electrochemical cell with the reaction shown below is 1.150 V. What is the Mn2+ concentration in this
cell?
Pb2*(aq) + Mn(s)– Pb(s) + Mn²+(aq)
[Mn2+] =
mol/L
Transcribed Image Text:When [Pb2+] = 1.40 M, the observed cell potential at 298 K for an electrochemical cell with the reaction shown below is 1.150 V. What is the Mn2+ concentration in this cell? Pb2*(aq) + Mn(s)– Pb(s) + Mn²+(aq) [Mn2+] = mol/L
Step 1
Decompose the cell reaction given into half-cell reactions and look up the standard electrode potentials.
Step 2
Use these to calculate the standard cell potential: E°cell = E°cathode - E°anode
Step 3
Determine the value of the reaction quotient, Q, using the value of E° cell and the measured value of Ecel-
At 298 K,
Ecel = Eel
0.0257 V
In Q
Step 4
Calculate the unknown concentration using the values of Q and the known concentrations.
Transcribed Image Text:Step 1 Decompose the cell reaction given into half-cell reactions and look up the standard electrode potentials. Step 2 Use these to calculate the standard cell potential: E°cell = E°cathode - E°anode Step 3 Determine the value of the reaction quotient, Q, using the value of E° cell and the measured value of Ecel- At 298 K, Ecel = Eel 0.0257 V In Q Step 4 Calculate the unknown concentration using the values of Q and the known concentrations.
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