Answer:
See explaination
Explanation:
1)
we know that
half cell with higher reduction potential is cathode
so
cathode :
N20 + 2H+ + 2e- ---> N2 + H20
anode :
Cr(s) ---> Cr+3 + 3e-
so
overall reaction is
3 N20 + 6H+ + 2 Cr ---> 3N2 + 3H20 + 2Cr+3
now
Eo cell = Eo cathode - Eo anode
so
EO cell = 1.77 + 0.74
Eo cell = 2.51 V
now
in this case
oxidizing agents are N20 and Cr+3
reducing agents are Cr and N2
higher the reduction potential , stronger the oxidizing agent
lower the reduction potential , stronger the reducing agent
so
oxidzing agents
N20 > Cr+3
reducing agents
Cr > N2
2)
cathode :
Au+ + e- --> Au
anode :
Cr ---> Cr+3 + 3e-
overall reaction
3Au+ + Cr ---> 3Au + Cr+3
Eo cell = 1.69 + 0.74
Eo cell = 2.43
now
oxidizing agents :
Au+ > Cr+3
reducing agents :
Cr > Au
3)
cathode :
N20 + 2H+ + 2e- ---> N2 + H20
andoe :
Au ---> Au+ + e-
overall
2 Au + N20 + 2H+ --> 2 Au+ + N2 + H20
Eo cell = 1.77 - 1.69
Eo cell = 0.08
oxidizing agents
N20 > Au+
reducing agents
Au > N2
<h3>Answer:</h3>
Correct Option-A (Ability to burn skin)
<h3>Explanation:</h3>
When skin tissues are exposed to Acids or Bases a chemical burn occurs as both of these substances are corrosive in nature. These burns occur without providing any heat, results from a very fast reaction, are extremely painful and causes damage to structures present under skin.
Option-B is incorrect because Acids taste sour, while, Bases taste bitter.
Option-C is incorrect because pH of Acids is less than 7 while, pH of Bases is greater than 7.
The pressure in the flask is 3.4 atm.
<em>pV</em> = <em>nRT
</em>
<em>T</em> = (20 + 273.15) K = 293.15 K
<em>p</em> = (<em>nRT</em>)/<em>V</em> = (1.4 mol × 0.082 06 L·atm·K⁻¹mol⁻¹ × 293.15 K)/10 L = 3.4 atm
Answer:
Explanation:
3Sn(NO3)2 (aq) + 2Cr(s) → 2Cr(NO3)3(aq ) + 3Sn(S )
The bottom of the food chain
