1903 Iowa Senate election

1903 Iowa Senate election

← 1901 November 3, 1903 1906[a] →
29 out of 50 seats in the Iowa State Senate 26 seats needed for a majority
Majority party Minority party   Party Republican Democratic Last election 39 11 Seats before 40[b] 10[b] Seats after 42 8 Seat change 2 2

Elections in Iowa
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v t e

The 1903 Iowa State Senate elections were the last state legislative general elections held on an odd-numbered year. Iowa voters elected state senators in 29 of the state senate's 50 districts. State senators traditionally serve four-year terms in the Iowa State Senate. However, under the Biennial Elections law enacted in 1904 by the Iowa General Assembly, the senators elected in 1903 served an additional fifth year (until the 1908 elections) to accommodate the transition to holding elections on even-numbered years.^[a]

A statewide map of the 50 state Senate districts in the 1903 elections is provided by the Iowa General Assembly here.

The 1903 elections occurred before primary elections were established in Iowa by the Primary Election Law in 1907.^[3] The general election took place on November 3, 1903.^[4]

Following the previous election, Republicans had control of the Iowa Senate with 39 seats to Democrats' 11 seats. However, during the twenty-ninth session of the Iowa General Assembly, the senators decided in March 1902 to decertify Democrat Joseph Martin Emmert of district eighteen and replace Emmert with Republican James E. Bruce, thus flipping the seat from Democratic to Republican control.^[b] Therefore, going into Election Day in 1903, Republicans held an advantage of 40 seats to Democrats' 10 seats.

To claim control of the chamber from Republicans, the Democrats needed to net 16 Senate seats.

Republicans maintained control of the Iowa State Senate following the 1903 general election with the balance of power shifting to Republicans holding 42 seats and Democrats having 8 seats (a net gain of 2 seats for Republicans).

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