1888 United States presidential election in Minnesota explained

See main article: 1888 United States presidential election.

Election Name:1888 United States presidential election in Minnesota
Country:Minnesota
Type:presidential
Ongoing:no
Previous Election:1884 United States presidential election in Minnesota
Previous Year:1884
Next Election:1892 United States presidential election in Minnesota
Next Year:1892
Election Date:November 6, 1888
Image1:Benjamin Harrison 1896.jpg
Nominee1:Benjamin Harrison
Party1:Republican Party (United States)
Home State1:Indiana
Running Mate1:Levi P. Morton
Electoral Vote1:7
Popular Vote1:142,192
Percentage1:54.12%
Nominee2:Grover Cleveland
Party2:Democratic Party (United States)
Home State2:New York
Running Mate2:Allen G. Thurman
Electoral Vote2:0
Popular Vote2:104,385
Percentage2:39.65%
Image3:Clinton B. Fisk drawing.png
Nominee3:Clinton Fisk
Party3:Prohibition Party
Home State3:New Jersey
Running Mate3:John A. Brooks
Electoral Vote3:0
Popular Vote3:15,511
Percentage3:5.82%
Map Size:350px
President
Before Election:Grover Cleveland
Before Party:Democratic Party (United States)
After Election:Benjamin Harrison
After Party:Republican Party (United States)
Flag Image:Flag of Minnesota (1893–1957).svg

The 1888 United States presidential election in Minnesota took place on November 6, 1888, as part of the 1888 United States presidential election. Voters chose seven representatives, or electors to the Electoral College, who voted for president and vice president.

Minnesota voted for the Republican nominee, Benjamin Harrison, over the Democratic nominee, incumbent President Grover Cleveland. Harrison won the state by a margin of 14.47%.

With 5.82% of the popular vote, Minnesota would be the Prohibition Party candidate Clinton Fisk’s strongest victory in terms of percentage in the popular vote.[1]

See also

Notes and References

  1. Web site: 1888 Presidential Election Statistics. Dave Leip’s Atlas of U.S. Presidential Elections. 2018-03-05.