Calculating the E of a galvanic cell So, I have $\ce{Cr2O7^2-(aq) + 14 H+(aq) + 6 I-(aq) -&gt; 2 Cr^3+(aq) + 3 I2(s) + 7 H2O(l)}$

and I know the concentrations to be like so:

$\ce{[Cr2O7^2-]}= 1.6\ \mathrm M$

$\ce{[H+]}= 2\ \mathrm M$

$\ce{[I-]}= 1.6\ \mathrm M$

$\ce{[Cr^3+]}= \pu{0.05271 M}$

And I want to calculate the $E(25\ \mathrm{^{\circ}C})$ of this galvanic process.

I know that I have to use $E=E^{\circ}-\frac{RT}{nF}\ln Q $, but I have a few questions:


How can I find $E^{\circ}$?
I know that $n$ represents the amount of electrons moles moving per mole of transformed material, but how can I find it when I have only the complete reaction like here?
With $Q=\ce{\frac{[Cr^{3+}]^2}{[H^+]^14[I^-]^6[Cr_2O_7^{2-}]}}$
I get $Q=\pu{6.32*10^{-9}V}$, which is very very small - does this seem to be the correct $Q$ for $\ln Q$?
Standard electrode potential can only be measured experimentally by comparing the half cells, in your case $\ce{Cr^{VI}/Cr^{III}}$ and $\ce{I2/I-}$ with a standard hydrogen electrode or a calomel electrode. The net standard electrode potential will be obtained by subtracting those of cathode-anode. You can find the standard electrode potential of most half cells in your textbook or online.
See in your reaction, two moles of $\ce{Cr^{VI}}$ turns to $\ce{Cr^{III}}$. So $n = 2 \times (6 - 3) = 6$. You can try out the same thing with the other half cell. The answer will be the same.