Is this correct? Thanks.
What is the standard cell potential?
Pb2+ ( aq ) + 2e „³ Pb (s) -0.13V
Mn2+ ( aq ) + 2e „³ Mn (s) -1.19V
Pb2+ ( aq ) + 2e „³ Pb (s) -0.13V
Mn (s) „³ Mn2+ ( aq ) + 2e ¡V +1.19V
The electrode potential for a given cell,
E = Ecathode ¡V Eanode = -0.13 ¡V(-1.19) = +1.06 V
I don't understand some of the symbols.
What is the standard cell potential?
Pb2+ ( aq ) + 2e „³ Pb (s) -0.13V
Mn2+ ( aq ) + 2e „³ Mn (s) -1.19V
Pb2+ ( aq ) + 2e „³ Pb (s) -0.13V
Mn (s) „³ Mn2+ ( aq ) + 2e ¡V +1.19V
The electrode potential for a given cell,
E = Ecathode ¡V Eanode = -0.13 ¡V(-1.19) = +1.06 V
There are some incorrect symbols in your text. Let me correct them for you:
What is the standard cell potential?
Pb2+ (aq) + 2e- → Pb(s) -0.13V
Mn2+ (aq) + 2e- → Mn(s) -1.19V
Pb2+ (aq) + 2e- → Pb(s) -0.13V
Mn(s) → Mn2+ (aq) + 2e- +1.19V
The electrode potential for a given cell,
E = E_cathode - E_anode = -0.13 - (-1.19) = +1.06 V.
In this context, the symbols used represent different chemical species and their respective electrode potentials. Here is an explanation of the symbols:
- Pb2+ (aq): This symbol represents lead ion in aqueous solution.
- 2e: This symbol represents two electrons being transferred in the chemical reaction.
- Pb (s): This symbol represents solid lead.
- Mn2+ (aq): This symbol represents manganese ion in aqueous solution.
- Mn (s): This symbol represents solid manganese.
- -0.13V: This value represents the electrode potential of the reaction where Pb2+ (aq) is reduced to Pb (s).
- -1.19V: This value represents the electrode potential of the reaction where Mn2+ (aq) is reduced to Mn (s).
When calculating the standard cell potential (E), you need to consider the cathode and anode reactions. The cathode is the site of reduction (where the species gains electrons), and the anode is the site of oxidation (where the species loses electrons).
Since the reduction potential for Pb2+ (aq) is -0.13V, it becomes the cathode reaction. The oxidation potential for Mn (s) is +1.19V, so it becomes the anode reaction.
To calculate the standard cell potential (E), you subtract the anode potential (Eanode) from the cathode potential (Ecathode). In this case, it is -0.13V - (-1.19V), which equals +1.06V.
Therefore, the standard cell potential for this reaction is +1.06V.