The dissociation of a/an ________ releases hydrogen ions and increases the concentration of hydrogen ions in a solution.

Calculate Ecell for the following electrochemical cell at 25 ºCPt (s) | H2 (g, 1.00 atm) | H+ (aq, 1.00 M) || Sn2+ (aq, 0.350 M)

katrin2010 [14]

Answer: The electrode potential of the cell is 0.093 V

Explanation:

The given chemical cell follows:

$Pt(s)|H_2(g,1atm)|H^+(aq,1.00M)||Sn^{4+}(aq,0.020M)|Sn^{2+}(aq.,0.350M)|Pt(s)$

Oxidation half reaction: $H_2(g,1atm)\rightarrow 2H^{+}(aq,1.0M)+2e^-;E^o_{2H^{+}/H_2}=0.0V$

Reduction half reaction: $Sn^{4+}(aq,0.020M)+2e^-\rightarrow Sn^{2+(aq.,0.350M);E^o_{Sn^{4+}/Sn^{2+}}=0.13V$

Net cell reaction:

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=0.13-(0.0)=0.13V$

To calculate the EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[H^{+}]^2[Sn^{2+}]}{[Sn^{4+}]}$

where,

$E_{cell}$ = electrode potential of the cell = ? V

$E^o_{cell}$ = standard electrode potential of the cell = +0.13 V

n = number of electrons exchanged = 2

$[H^{+}]=1.00M$

$[Sn^{2+}]=0.350M$

$[Sn^{4+}]=0.020M$

Putting values in above equation, we get:

$E_{cell}=0.13-\frac{0.059}{2}\times \log(\frac{(1.0)^2\times 0.350}{0.020})$

$E_{cell}=0.093V$

Hence, the electrode potential of the cell is 0.093 V

3 0

4 years ago

What is true about the electrolysis of water? Use the picture to choose 2 correct answers.

Ainat [17]

Answer:

oxygen is produced at the anode and hydrogen gas is produced at the cathode

Explanation:

7 0

4 years ago

Need help please?!!!!!

puteri [66]

(a) 33.6 L of oxygen would be produced.

(b) 106 grams of $Na_2CO_3$ would be needed

<h3>Stoichiometric calculations</h3>

1 mole of gas = 22.4 L

(a) From the equation, 2 moles of $KClO_3$ produce 3 moles of $O_2$ . 1 mole of $KClO_3$ will, therefore, produce 1.5 moles of $O_2$ .

1.5 moles of oxygen = 22.4 x 1.5 = 33.6 L

(b) 22.4 L of $CO_2$ is produced at STP. This means that 1 mole of the gas is produced.

From the equation, 1 mole of $CO_2$ requires 1 mole of $Na_2CO_3$ .

Molar mass of $Na_2CO_3$ = (23x2)+ (12)+(16x3) = 106 g/mol

Mass of 1 mole $Na_2CO_3$ = 1 x 106 = 106 grams

More on stoichiometric calculations can be found here: brainly.com/question/27287858

#SPJ1

6 0

2 years ago

Please please help me

NNADVOKAT [17]

Answer:hey man hey man

Explanation:

ooo she fine

4 0

3 years ago

Can ya help me out real quick?

Taya2010 [7]

Answer:

302 kj heat is released by lowering the temperature

8 0

3 years ago