1892 United States presidential election in Wisconsin

The 1892 United States presidential election in Wisconsin was held on November 8, 1892 as part of the 1892 United States presidential election. Wisconsin voters chose twelve electors to the Electoral College, who voted for president and vice president.

1892 United States presidential election in Wisconsin

← 1888 November 8, 1892 1896 →
  Grover Cleveland - NARA - 518139.tif Pach Brothers - Benjamin Harrison.jpg
Nominee Grover Cleveland Benjamin Harrison
Party Democratic Republican
Home state New York Indiana
Running mate Adlai Stevenson I Whitelaw Reid
Electoral vote 12 0
Popular vote 177,325 171,101
Percentage 47.72% 46.05%

President before election

Benjamin Harrison
Republican

Elected President

Grover Cleveland
Democratic

Democratic Party candidate Grover Cleveland won Wisconsin with 48% of the popular vote, winning the state's twelve electoral votes.[1] As a result of his win, Cleveland became the first Democratic presidential candidate since Franklin Pierce in 1852 to win Wisconsin. No Democrat would win Wisconsin again until Woodrow Wilson in 1912.

ResultsEdit

1892 United States presidential election in Wisconsin
Party Candidate Votes Percentage Electoral votes
Democratic Grover Cleveland 177,325 47.72% 12
Republican Benjamin Harrison (incumbent) 171,101 46.05% 0
Prohibition John Bidwell 13,136 3.54% 0
People's James Weaver 10,019 2.7% 0
Totals 371,581 100.0% 12

ReferencesEdit

  1. ^ "1892 Presidential General Election Results - Wisconsin". Retrieved August 18, 2016.