Enter the balanced chemical equation including states that describes the electrochemical cell that is represented by the cell notation as shown.
Pt(s) l H2O(l) l O2(g) l H+(aq) ll H+(aq), Cr2O72-(aq), Cr3+(aq) l Pt(s)
2H2O ==> O2 + 4H^+ 4e
H^+ + Cr2O7^-2 + 6e ==> 2Cr^+3 + 7H2O
You can add the phases. Check my work.
Okay so the problem I had with this is like...I did't understand that when it was written say on the left of the double vertical line how you know
H2O --> O2 + H+
will it always be written in the order of how you write the reaction? or like could you also right it like
O2 + H+ --> h2o
or basically how do you know which species (H2), O2, and H+ are on the reactant/product side?
also shouldn't it be the coefficient of H^+ in the second equation be 14?
To write the balanced chemical equation for the electrochemical cell represented by the given cell notation, we need to understand the components of the cell notation.
The cell notation consists of two half-cells, separated by a double vertical line (||). The anode half-cell is written to the left of the double vertical line, and the cathode half-cell is written to the right.
Let's break down the given cell notation:
Pt(s) | H2O(l) | O2(g) | H+(aq) || H+(aq), Cr2O72-(aq), Cr3+(aq) | Pt(s)
The anode half-cell is Pt(s) | H2O(l) | O2(g) | H+(aq), and the cathode half-cell is H+(aq), Cr2O72-(aq), Cr3+(aq) | Pt(s).
Now, we can write the balanced chemical equations for these half-cells individually.
1. Anode half-cell:
The anode half-cell consists of the oxidation of H2O to O2. The balanced equation for this process is:
2H2O(l) -> O2(g) + 4H+(aq) + 4e-
2. Cathode half-cell:
The cathode half-cell involves the reduction of Cr2O72- to Cr3+. The balanced equation for this process is:
Cr2O72-(aq) + 14H+(aq) + 6e- -> 2Cr3+(aq) + 7H2O(l)
To balance the overall reaction, we need to multiply all the species in the equations by appropriate coefficients so that the number of electrons transferred in both half-reactions is equal.
By multiplying the anode half-cell by 6 and the cathode half-cell by 4, we obtain:
6(2H2O(l) -> O2(g) + 4H+(aq) + 4e-) (anode)
4(Cr2O72-(aq) + 14H+(aq) + 6e- -> 2Cr3+(aq) + 7H2O(l)) (cathode)
After multiplying and simplifying, the balanced overall equation can be written as:
6H2O(l) + 2Cr2O72-(aq) + 14H+(aq) -> O2(g) + 4Cr3+(aq) + 14H2O(l)
Simplifying further:
6H2O(l) + 2Cr2O72-(aq) + 14H+(aq) -> O2(g) + 4Cr3+(aq) + 14H2O(l)
Therefore, the balanced chemical equation for the given electrochemical cell notation is:
6H2O(l) + 2Cr2O72-(aq) + 14H+(aq) -> O2(g) + 4Cr3+(aq) + 14H2O(l)