Write the standard line notation for each cell below.
IO3-(aq) + Fe2+(aq)--> Fe3+(aq) +I2(aq)
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H2O2 + 2 H+ + 2 e- --> 2 H2O E = 1.78 V
O2 + 2 H+ + 2 e- --> H2O2 E= 0.68 V
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Mn2+ + 2 e- --> Mn E = -1.18 V
Fe3+ + 3 e- --> Fe E = -0.036 V
Oh, chemical reactions, how electrifying! Let me brighten things up with some clownish humor while I jot down the standard line notation for each cell:
1. IO3-(aq) + Fe2+(aq) --> Fe3+(aq) + I2(aq)
Standard line notation: Fe2+(aq) | Fe3+(aq) || IO3-(aq) | I2(aq)
2. H2O2 + 2 H+ + 2 e- --> 2 H2O E = 1.78 V
Standard line notation: Pt | H2O2 | 2H+(aq) || 2H2O
3. O2 + 2 H+ + 2 e- --> H2O2 E = 0.68 V
Standard line notation: Pt | O2 | 2H+(aq) || H2O2
4. Mn2+ + 2 e- --> Mn E = -1.18 V
Standard line notation: Mn2+(aq) | Mn
5. Fe3+ + 3 e- --> Fe E = -0.036 V
Standard line notation: Fe3+(aq) | Fe
Hope these electrifying jokes and standard line notation put a spark of laughter in your chemistry mind!
To write the standard line notation for each cell, we need to identify the oxidation and reduction half-reactions and indicate the species involved in each reaction.
1. IO3-(aq) + 6 H+ + 6 e- --> 3 I2(aq) + 3 H2O
Fe2+(aq) --> Fe3+(aq) + 1 e-
Standard line notation:
Fe2+(aq) | Fe3+(aq) || IO3-(aq) | I2(aq)
2. H2O2 + 2 H+ + 2 e- --> 2 H2O
O2 + 2 H+ + 2 e- --> H2O2
Standard line notation:
O2(g) | H2O2(aq) || H2O2(aq) | H2O(l)
3. Mn2+ + 2 e- --> Mn
Fe3+ + 3 e- --> Fe
Standard line notation:
Fe3+(aq) | Fe(s) || Mn2+(aq) | Mn(s)
To write the standard line notation for each cell, we need to follow a specific format. The standard line notation consists of three components: the oxidation half-reaction, the reduction half-reaction, and the overall cell reaction.
Let's break it down for each cell:
1. IO3-(aq) + Fe2+(aq) --> Fe3+(aq) + I2(aq)
Oxidation Half-reaction: IO3-(aq) --> I2(aq)
Reduction Half-reaction: Fe2+(aq) --> Fe3+(aq)
To balance the number of electrons on each side, the oxidation half-reaction needs to be multiplied by 6 and the reduction half-reaction by 2:
6 IO3-(aq) --> 6 I2(aq)
2 Fe2+(aq) --> 2 Fe3+(aq)
Overall Cell Reaction: Adding the oxidation and reduction half-reactions:
6 IO3-(aq) + 2 Fe2+(aq) --> 6 I2(aq) + 2 Fe3+(aq)
2. H2O2 + 2 H+ + 2 e- --> 2 H2O E = 1.78 V
O2 + 2 H+ + 2 e- --> H2O2 E= 0.68 V
The equation with the higher voltage is the reduction half-reaction, so let's assign that to:
Reduction Half-reaction: O2 + 2 H+ + 2 e- --> H2O2
To balance the number of electrons, we can multiply the oxidation half-reaction by 2:
2 H2O2 --> 4 H+ + 4 e-
Overall Cell Reaction: Adding the oxidation and reduction half-reactions:
O2 + 2 H+ + 2 e- + 2 H2O2 --> H2O2 + 4 H+ + 4 e-
Simplifying the equation by canceling out common species:
O2 + 2 H2O2 --> 2 H2O
3. Mn2+ + 2 e- --> Mn E = -1.18 V
Fe3+ + 3 e- --> Fe E = -0.036 V
The equation with the higher voltage is the reduction half-reaction, so let's assign that to:
Reduction Half-reaction: Fe3+ + 3 e- --> Fe
To balance the number of electrons, we can multiply the oxidation half-reaction by 3:
3 Mn2+ --> 3 Mn + 6 e-
Overall Cell Reaction: Adding the oxidation and reduction half-reactions:
3 Mn2+ + Fe3+ + 3 e- --> 3 Mn + Fe
Note that there is no need to balance the overall cell reaction further because the electrons are already balanced.
Did you mean I2(s)? I think I2 usually is a solid so this would be a phase boundry and shown as I've done below. The () is where concns would go. You show aq there. You can write aq or the concn there. Here is a good site for reading.
http://chemed.chem.wisc.edu/chempaths/GenChem-Textbook/Cell-Notation-and-Conventions-995.html
Pt, IO3^-(), Fe^2+()||Fe^3+()|I2, Pt