# Problem: Consider the following half-reactions and their standard reduction potentials then give the standard line (cell) notation for a voltaic cell built on these half reactions.Mn2+(aq) + 2 e- &lt;=&gt; Mn(s) E° =  -1.18 VFe3+(aq) + 3 e- &lt;=&gt; Fe(s) E° =  -0.036 Va. Mn2+(aq, 1.0 M) | Mn(s) || Fe(s) | Fe3+(aq, 1.0 M)b. Mn(s) | Mn2+(aq, 1.0 M) || Fe3+(aq, 1.0 M) | Fe(s)c. Fe3+(aq, 1.0 M) | Fe(s) || Mn(s) | Mn2+(aq, 1.0 M)d.  Fe(s) | Fe3+(aq, 1.0 M) ||  Mn2+(aq, 1.0 M) | Mn(s)e.  Mn2+(aq, 1.0 M) | Fe3+(aq, 1.0 M) ||  Fe(s) | Mn(s)

###### FREE Expert Solution

We’re being asked to give the standard line (cell) notation for a voltaic cell built on these half reactions below:

Mn2+(aq) + 2 e- ⇌ Mn(s)           E° =  -1.18 V
Fe3+(aq) + 3 e- ⇌ Fe(s)             E° =  -0.036 V

When writing a cell notation, we use the following format – “as easy as ABC To determine the cell notation of the reaction, we will use the following steps:

Step 1: Determine the anode and the cathode.
Step 2: Write the cell notation of the reaction. ###### Problem Details

Consider the following half-reactions and their standard reduction potentials then give the standard line (cell) notation for a voltaic cell built on these half reactions.

Mn2+(aq) + 2 e- <=> Mn(s) E° =  -1.18 V

Fe3+(aq) + 3 e- <=> Fe(s) E° =  -0.036 V

a. Mn2+(aq, 1.0 M) | Mn(s) || Fe(s) | Fe3+(aq, 1.0 M)
b. Mn(s) | Mn2+(aq, 1.0 M) || Fe3+(aq, 1.0 M) | Fe(s)
c. Fe3+(aq, 1.0 M) | Fe(s) || Mn(s) | Mn2+(aq, 1.0 M)
d.  Fe(s) | Fe3+(aq, 1.0 M) ||  Mn2+(aq, 1.0 M) | Mn(s)
e.  Mn2+(aq, 1.0 M) | Fe3+(aq, 1.0 M) ||  Fe(s) | Mn(s)