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2 years ago

12. A voltaic cell consists of a chromium electrode dipped in a 1.20 M chromium (III) nitrate

1 answer:

5 0

For a voltaic cell consisting of chromium, an electrode dipped in a 1.20 M chromium (III) nitrate solution and a tin electrode dipped in a 0.400 M tin (II) nitrate solution, the cell potential at 298 K is mathematically given as

Ecell = 0.577 V

<h3 /><h3>What is the cell potential at 298 K?</h3>

Generally, the equation for the Oxidation and Reduction is mathematically given as

Cr(s) ------------------ Cr+3(aq) + 3e- ] x 2 ...O

Sn+2(aq) + 2e- ------------ Sn(s) ] x 3 ...R

Reaction

2 Cr(s) + 3 Sn+2(aq) --------------- 2 Cr+3(aq) + 3 Sn(s)

Therefore

Eicell = - 0.14 - ( - 0.74)

Eicell = 0.60

In conclusion

$Ecell= E0cell - \frac{0.0591}{n} * \frac{log[Cr+3]^2}{ [ Sn+2]^3}$

$Ecell = 0.60 - \frac{0.0591 }{6} \frac{log( 1.20)^2}{ ( 0.200)^3}$

Ecell = 0.577 V

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Explanation:

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D. As white light passes through a prism , it bends and separates into different colors

Explanation:

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So, the correct answer is

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In the gas-phase reaction 2 A + B ⇌ 3 C + 2 D, it was found that, when 1.00 mol A, 2.00 mol B, and 1.00 mol D were mixed and all

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Answer:

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$B=\frac{1.4}{4.3} \\=0.3256$

$C=\frac{0.9}{4.3} \\=0.2093$

$D=\frac{1.6}{4.3} \\=0.3721$

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Explanation:

The given equation is :

$2A+B$ ⇄ $3C+2D$

Let $\alpha$ be the number of moles dissociated per mole of $B$

Thus ,

<em>The initial number of moles of :</em>

$A=1$

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$C=0$

$D=1$

$2A\\(1-2\alpha)$ + $B\\2(1-\alpha)$ ⇄ $3C\\(3\alpha)$ + $2D\\(1+2\alpha)$

And finally the number of moles of $C[tex] is 0.9Thus ,[tex]3\alpha=0.9\\\alpha=0.3[tex]The final number of moles of:[tex]A = 1-2\alpha=1-2*0.3=0.4mol[tex] [tex]B=2(1-\alpha)=2(1-0.3)=1.4mol[tex][tex]D=1+2\alpha=1+2*0.3=1.6mol[tex]Thus , total number of moles are : 0.4+1.4+0.9+1.6=4.3(i)The mole fractions are : [tex]A=\frac{0.4}{4.3} \\=0.0930$